Probability - Class 11 Mathematics - Chapter 14 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Probability | NCERT | Mathematics | Class 11
Two coins are tossed simultaneously 200 times and the number of heads is recorded as follows:
\begin{tabular}{|c|c|} \hline Number of heads & Frequency \ \hline 0 & 65 \ \hline 1 & 85 \ \hline 2 & 50 \ \hline \end{tabular}
What is the probability of 1 head appearing?
A) 0.4 B) 0.5 C) 0.425 D) 0.75
Answer: C) 0.425
To determine the probability of 1 head appearing, we use the formula for probability $P(E)$ where $E$ is the event of interest:
$$ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of trials}} $$
In this scenario, the event $E$ (1 head appearing) occurred 85 times out of 200 total coin tosses. Applying the formula, we get:
$$ P(E) = \frac{85}{200} = 0.425 $$
Thus, the probability that exactly one head appears when tossing two coins simultaneously and observing the outcomes across 200 tosses is 0.425.
"A four-digit number (ranging from 0000 to 9999) is said to be lucky if the sum of its first two digits is equal to the sum of its last two digits. If a four-digit number is picked at random, then the probability that it is a lucky number is:
A. 0.07
B. 0.067
C. 0.67
D. 0.6"
Solution
The correct answer is B. 0.067.
The total number of four-digit numbers ranging from $0000$ to $9999$ is $10^4 = 10000$.
For a four-digit number to be considered lucky, the sum of the first two digits should equal the sum of the last two digits. Suppose this sum equals $k$. The task is to determine the number of solutions to the equation $x_1 + x_2 = k$ where $0 \leq k \leq 18$ and $x_i$ are the individual digits.
The solutions to these equations are represented by the coefficient of $x^k$ in the expansion of $(1 + x + x^2 + \ldots + x^9)^2$. This is because each power represents a possible sum of the digits from $0$ to $9$.
Thus, the probability that a randomly chosen number is lucky can be calculated as: $$ \frac{\sum_{k=0}^{18} a_k^2}{10000} $$ where $a_k$ is the coefficient of $x^k$ in the expansion.
The generating function for the solutions for each $k$ is given by: $$ (1+x+x^2+\ldots+x^9)^2 = a_0 + a_1x + a_2x^2 + \ldots + a_{18}x^{18} $$
To find $\sum_{k=0}^{18} a_k^2$, we evaluate the constant term in: $$ (a_0 + a_1x + \ldots + a_{18}x^{18})(a_0 + \frac{a_1}{x} + \ldots + \frac{a_{18}}{x^{18}}) $$ which simplifies to: $$ \text{Coefficient of } x^{18} \text{ in } (1-x^{10})^4(1-x)^{-4} $$
This further simplifies to: $$ 21C_3 - 4 \times 11C_3 = 1330 - 660 = 670 $$
Hence, the required probability that the four-digit number is lucky is: $$ \frac{670}{10000} = 0.067 $$
"A die and a coin are tossed simultaneously. If a head on the coin is taken to be 1 and a tail is taken to be 0, find the number of outcomes where the sum of the numbers on the die and the coin is greater than 4."
A) 2
B) 3
C) 5
D) 6
The correct option is C) 5
When a die and a coin are tossed simultaneously, the possible outcomes for the die range from 1 through 6 (inclusive) and the outcomes for the coin are head ($H=1$) and tail ($T=0$). Thus, the total outcomes generated from one toss can be described as pairs of numbers: $(\text{die outcome}, \text{coin outcome})$.
The complete list of possible outcomes is: $$ (1,0), (1,1), (2,0), (2,1), (3,0), (3,1), (4,0), (4,1), (5,0), (5,1), (6,0), (6,1) $$
To find outcomes where the sum of the numbers on the die and the coin is greater than 4, we need to calculate the sum in each case:
- $(1,0) = 1$
- $(1,1) = 2$
- $(2,0) = 2$
- $(2,1) = 3$
- $(3,0) = 3$
- $(3,1) = 4$
- $(4,0) = 4$
- $(4,1) = 5$
- $(5,0) = 5$
- $(5,1) = 6$
- $(6,0) = 6$
- $(6,1) = 7$
The favorable outcomes that result in sums greater than 4 are: $$ (4,1), (5,0), (5,1), (6,0), (6,1) $$
Hence, the number of outcomes is 5.
Getting a number less than 7 while rolling a dice is not a sure event.
A) True
B) False
The correct option is B False
On a standard die, all the face values (1, 2, 3, 4, 5, 6) are indeed less than 7. Thus, the number of favorable outcomes when trying to roll less than 7 is 6.
The probability ( P ) of rolling a number less than 7 is calculated as: $$ P = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} = \frac{6}{6} = 1 $$
An event with probability 1 is termed a sure event. Therefore, rolling a number less than 7 with a die is definitely a sure event.
Football teams $T_{1}$ and $T_{2}$ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of $T_{1}$ winning, drawing, and losing a game against $T_{2}$ are $\frac{1}{2}, \frac{1}{6}$ and $\frac{1}{3}$, respectively. Each team gets 3 points for a win, 1 point for a draw, and 0 points for a loss in a game. Let $X$ and $Y$ denote the total points scored by teams $T_{1}$ and $T_{2}$, respectively, after two games.
$P(X=Y)$ is
A) $\frac{11}{36}$
B) $\frac{1}{3}$
C) $\frac{13}{36}$
D) $\frac{1}{2}$
The problem requires us to find $P(X = Y)$, which is the probability that teams $T_{1}$ and $T_{2}$ score an equal total of points after two games.
Given the probabilities:
- $T_{1}$ winning a game: $\frac{1}{2}$
- $T_{1}$ drawing a game: $\frac{1}{6}$
- $T_{1}$ losing a game: $\frac{1}{3}$ (since probabilities sum to 1).
For equal points, there are two scenarios:
- Both teams winning one game each: This can happen if $T_{1}$ wins the first and $T_{2}$ wins the second, or $T_{2}$ wins the first and $T_{1}$ wins the second.
- Both games result in draws.
The probabilities for these scenarios are:
- $T_{1}$ wins first and $T_{2}$ wins second: $\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$
- $T_{2}$ wins first and $T_{1}$ wins second: $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$
- Both games are draws: $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$
The total probability of these scenarios is: $$ P(X = Y) = \left(\frac{1}{6} + \frac{1}{6}\right) + \left(\frac{1}{36}\right) = \frac{1}{3} + \frac{1}{36} = \frac{12}{36} + \frac{1}{36} = \frac{13}{36} $$
Thus, $P(X = Y) = \frac{13}{36}$, making the correct option C.
An urn contains 6 red, 4 blue, and 3 green marbles. If 2 marbles are picked at random, then what is the probability that both marbles are red?
(A) $21 / 26$
(B) $5 / 26$
(C) $10 / 13$
(D) $3 / 13$
(E) None of these
The correct answer is (B) $5 / 26$.
To find the probability that both marbles picked are red, we first determine the total number of ways to choose 2 marbles from the 13 available (6 red + 4 blue + 3 green). This can be calculated by the combination formula $C(n, k)$ which represents choosing $k$ items from $n$ without regard to order. Thus, the total number of ways to select 2 marbles from 13 is:
$$ \binom{13}{2} = \frac{13 \times 12}{2 \times 1} = 78 $$
Next, we calculate the number of ways to specifically pick 2 red marbles out of the 6 available. This is:
$$ \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 $$
The probability that both marbles are red is then given by the ratio of the number of favorable outcomes (selecting 2 red marbles) to the total number of outcomes (selecting any 2 marbles):
$$ P(\text{Both marbles are red}) = \frac{15}{78} = \frac{5}{26} $$
Thus, the probability that both marbles are red is $5/26$.
In a deck of 52 cards, what is the probability of getting: (i) an even number card? (ii) a black ace card?
Solution
Total number of cards: $52$
Part (i): Probability of getting an even number card
- Even numbered cards in one suit: $5$
Since there are 4 suits, the total number of even cards is: $$ 5 \times 4 = 20 $$ The probability of drawing an even number card: $$ P(\text{{getting an even number card}}) = \frac{20}{52} = \frac{5}{13} $$
Part (ii): Probability of getting a black ace card
- Number of black aces: $2$ (one Ace of Spades and one Ace of Clubs)
The probability of drawing a black ace: $$ P(\text{{getting a black ace card}}) = \frac{2}{52} = \frac{1}{26}
A die is rolled twice. Let us define the following events:
A: "Getting doublets"
B: "Getting a total of at least 11"
How many outcomes are common between $A$ and $B$?
A. One
B. Two
C. Three
D. Four
The correct answer is A. One.
Analysis
-
Event A (Getting doublets): consists of outcomes where both dice show the same number. The sample space for A is: $$ A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} $$
-
Event B (Getting a total of at least 11): includes outcomes where the sum of numbers on both dice is at least 11. The sample space for B is: $$ B = {(5, 6), (6, 6), (6, 5)} $$
Intersection of A and B
To find outcomes common to both $A$ and $B$, we look for intersections in these sample spaces: $$ A \cap B = {(6, 6)} $$ There is precisely one common outcome between event $A$ and event $B$, which is the outcome $(6, 6)$.
Thus, the number of outcomes common between $A$ and $B$ is one.
Two coins are tossed together. Find the probability of getting: (i) exactly one tail (ii) at least one head (iii) no head (iv) at most one head
Here's the solution structured in a clearer and more detailed way:
- Total number of outcomes: When two coins are tossed, there are four possible outcomes: $$ {\text{HH}, \text{HT}, \text{TT}, \text{TH}} $$
(i) Finding the probability of getting exactly one tail:
- Favorable outcomes: HT and TH
- Total favorable outcomes: $$ 2 $$
- Probability (P) is calculated as: $$ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{4} = \frac{1}{2} $$
(ii) Finding the probability of getting at least one head:
- Favorable outcomes: HH, HT, TH (all outcomes except TT)
- Total favorable outcomes: $$ 3 $$
- Probability (P) of getting at least one head: $$ P(E) = \frac{3}{4} $$
(iii) Finding the probability of getting no head:
- Favorable outcomes: TT
- Total favorable outcomes: $$ 1 $$
- Probability (P) of getting no head: $$ P(E) = \frac{1}{4} $$
(iv) Finding the probability of getting at most one head:
- Favorable outcomes: HT, TH, TT (all outcomes except HH)
- Total favorable outcomes: $$ 3 $$
- Probability (P) of getting at most one head: $$ P(E) = \frac{3}{4} $$
In summary:
- The probability of getting exactly one tail is $$ \frac{1}{2} $$.
- The probability of getting at least one head is $$ \frac{3}{4} $$.
- The probability of getting no head is $$ \frac{1}{4} $$.
- The probability of getting at most one head is also $$ \frac{3}{4} $$.
The last page number on the textbooks of different subjects in a class is given. \begin{tabular}{|c|c|} \hline Subject & Page Number \ \hline English & 203 \ \hline Hindi & 203 \ \hline Maths & 152 \ \hline Physics & 177 \ \hline Chemistry & 225 \ \hline Biology & 172 \ \hline Computer Sci. & 185 \ \hline History & 203 \ \hline Geography & 172 \ \hline Moral Sci. & 267 \ \hline \end{tabular}
Find the probability that the last page number ends in (i) an odd number (ii) an even number (iii) a multiple of 3. [4 MARKS]
Solution
Concept Explanation: To find the probability of certain characteristics of page numbers (odd, even, multiple of 3), we calculate the proportion of books that meet the criteria out of the total number of books.
Total Number of Books: $$ 10 $$ (1 Mark for setting up the problem)
(i) Probability of last page number ending in an odd number:
- Books with page numbers ending in odd numbers: $7$
- Required Probability: $$ \frac{7}{10} = 0.7 $$ (1 Mark for calculation)
(ii) Probability of last page number ending in an even number:
- Books with page numbers ending in even numbers: $3$
- Required Probability: $$ \frac{3}{10} = 0.3 $$ (1 Mark for calculation)
(iii) Probability of last page number ending as a multiple of 3:
- Books with page numbers that are multiples of 3: $3$
- Required Probability: $$ \frac{3}{10} = 0.3 $$ (1 Mark for calculation)
This approach uses a straightforward counting technique to evaluate each probability based on the given data.
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is: (i) a black king (ii) either a black card or a king (iii) black and a king (iv) a jack, queen, or a king (v) neither a heart nor a king (vi) spade or an ace (vii) neither a red card nor a queen (ix) other than an ace (x) a ten (xi) a spade (xii) a black card (xiii) the seven of clubs (xiv) jack (xv) the ace of spades (xvi) a queen (xvii) a heart (xviii) a red card
Based on the question to determine the probabilities for various scenarios with a deck of 52 cards, here are the explained probabilities:
- Total number of possible outcomes for any event drawn from the standard deck of 52 cards is $52$.
-
Probability of drawing a black king: There are only 2 black kings (King of Clubs and King of Spades). $$ P(\text{Black King}) = \frac{2}{52} = \frac{1}{26} $$
-
Probability of drawing either a black card or a king: There are 26 black cards and 2 additional kings that are red. $$ P(\text{Black card or King}) = \frac{26 + 2}{52} = \frac{28}{52} = \frac{7}{13} $$
-
Probability of drawing a black king: This scenario is the same as the first, with only 2 black kings. $$ P(\text{Black and King}) = \frac{2}{52} = \frac{1}{26} $$
-
Probability of drawing a jack, queen, or king: Each rank (jack, queen, king) has 4 cards, totaling 12 cards. $$ P(\text{Jack, Queen, or King}) = \frac{12}{52} = \frac{3}{13} $$
-
Probability of drawing neither a heart nor a king: Calculate total hearts and kings then subtract overlaps. $$ P(\text{Neither Heart nor King}) = \frac{39 - 3}{52} = \frac{36}{52} = \frac{9}{13} $$
-
Probability of drawing a spade or an ace: There are 13 spades, and an additional 3 aces from other suits. $$ P(\text{Spade or Ace}) = \frac{16}{52} = \frac{4}{13} $$
-
Probability of drawing neither a red card nor a queen: With 26 red cards minus the red queens. $$ P(\text{Neither Red Card nor Queen}) = \frac{52 - 26 - 2}{52} = \frac{24}{52} = \frac{6}{13} $$
-
Probability of drawing something other than an ace: There are 4 aces in a deck of 52 cards. $$ P(\text{Other than Ace}) = \frac{48}{52} = \frac{12}{13} $$
-
Probability of drawing a ten: There are 4 tens in a deck. $$ P(\text{Ten}) = \frac{4}{52} = \frac{1}{13} $$
-
Probability of drawing a spade: There are 13 spades. $$ P(\text{Spade}) = \frac{13}{52} = \frac{1}{4} $$
-
Probability of drawing any black card: There are 26 black cards. $$ P(\text{Black Card}) = \frac{26}{52} = \frac{1}{2} $$
-
Probability of drawing the seven of clubs: There is only one seven of clubs. $$ P(\text{Seven of Clubs}) = \frac{1}{52} $$
-
Probability of drawing a jack: There are 4 jacks. $$ P(\text{Jack}) = \frac{4}{52} = \frac{1}{13} $$
-
Probability of drawing the ace of spades: There is one ace of spades. $$ P(\text{Ace of Spades}) = \frac{1}{52} $$
-
Probability of drawing a queen: There are 4 queens. $$ P(\text{Queen}) = \frac{4}{52} = \frac{1}{13} $$
-
Probability of drawing a heart card: There are 13 hearts. $$ P(\text{Heart}) = \frac{13}{52} = \frac{1}{4} $$
-
Probability of drawing a red card: There are 26 red cards. $$ P(\text{Red Card}) = \frac{26}{52} = \frac{1}{2} $$
Each of these probabilities provides clear insight into drawing specific outcomes from a standard deck of 52 cards, identifying factors that affect the probability of different types of draws.
A fair dice is thrown eight times. What is the probability that a third 6 is obtained on the eighth throw?
$\begin{array}{l}{ }^{7} C_{2}\left(\frac{5^6}{6^8}\right) \ { }^{7} C_{2}\left(\frac{5^6}{6^2}\right) \ { }^{7} C_{2}\left(\frac{5}{6}\right) \ { }^{7} C_{2}\left(\frac{5\cdot 6}{6^2}\right)\end{array}$
The correct answer is Option D: $$ { }^{7} C_{2}\left(\frac{5^5}{6^8}\right) $$
To find the probability that the third occurrence of '6' happens on the eighth throw, the first step is to ensure that two '6's occur within the first seven throws. Next, a '6' must occur specifically on the eighth throw.
The probability can thus be computed by considering:
- Choosing 2 throws out of the first 7 to be '6's: $ { }^{7}C_{2} $.
- Each '6' in these chosen throws has a probability of $ \frac{1}{6} $, and non-'6' outcomes (1 to 5) in the other throws have a probability of $ \frac{5}{6} $.
- Finally, getting a '6' on the eighth throw also has a probability of $ \frac{1}{6} $.
Assembling this information, the full probability calculation is: $$ { }^{7}C_{2} \times \left(\frac{1}{6}\right)^2 \times \left(\frac{5}{6}\right)^5 \times \frac{1}{6} = { }^{7}C_{2}\left(\frac{5^5}{6^8}\right) $$ Thus, the probability that the third '6' is obtained on the eighth throw is given by ${ }^{7} C_{2}\left(\frac{5^5}{6^8}\right)$, highlighting Option D as the correct choice.
If $p$ is the probability that a man aged $x$ years will die in a year, find the probability that out of $n$ men $A_{1}, A_{2}, \ldots, A_{n}$, each aged $x$ years, $A_{1}$ will die in a year and will be the first to die.
A) $= \left(\frac{1}{n}\right)\left[1-p^{n}\right]$.
B) $= \left(\frac{1}{n}\right)\left[1-(1-p)^{n}\right]$.
C) $= \left(\frac{1}{n}\right)\left[(1-p)^{n}\right]$.
D) $= \left(\frac{1}{n}\right)\left[1-(2-p)^{n}\right]$.
The correct option is B $$ = \left(\frac{1}{n}\right)\left[1 - (1-p)^n\right]. $$
Given that the probability that a man aged $x$ years will die within a year is $p$, the probability that he does not die within a year is $1 - p$. Thus, the probability that none of the $n$ men, all aged $x$ years, will die within the year is $(1-p)^n$.
We're also interested in the event that at least one man among the $n$ men dies within the year. This probability is given by $1 - (1-p)^n$.
Since any of the $n$ men is equally likely to be the first to die if one dies, each has a probability of $\frac{1}{n}$ of being the first.
Therefore, the probability that $A_1$ both dies within the year and is the first to die among the $n$ men is: $$ = \left(\frac{1}{n}\right)\left[1 - (1-p)^n\right]. $$
A bag contains $(2n + 1)$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $(n + 1)$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$, then $n$ is equal to:
A) 10
B) 11
C) 12
D) 13
The correct choice is A) 10. Here is a detailed explanation of how to arrive at that answer:
The bag contains coins that are biased (with heads on both sides) and fair coins (with one head and one tail). Since there are $n$ double-headed coins and $(n+1)$ fair coins, the total number of coins is $2n + 1$.
Calculating the probability of getting heads when picking and tossing one coin randomly involves:
-
Choosing a double-headed coin - This occurs with probability $$\frac{n}{2n+1}$$ because there are $n$ such coins out of $2n + 1$. These coins will always land on heads, so the probability of getting heads after picking such a coin is 1.
-
Choosing a fair coin - This happens with a probability $$\frac{n+1}{2n+1}$$ because there are $n+1$ fair coins out of $2n + 1$. A fair coin lands on heads with a probability of $\frac{1}{2}$.
So, the total probability of obtaining heads $P(\text{head})$ is: $$ P(\text{head}) = \left(\frac{n}{2n+1}\right)(1) + \left(\frac{n+1}{2n+1}\right)\left(\frac{1}{2}\right) = \frac{n}{2n+1} + \frac{n+1}{2(2n+1)} $$
According to the problem, $P(\text{head}) = \frac{31}{42}$. Thus, we set up an equation: $$ \frac{n}{2n+1} + \frac{n+1}{2(2n+1)} = \frac{31}{42} $$
Solving the equation by clearing the denominators: $$ 2n + \frac{n+1}{2} = (2n+1)\left(\frac{31}{42}\right) $$
To remove the fraction, multiply both sides by 42: $$ 84n + 42\left(\frac{n+1}{2}\right) = 62(2n+1) $$
Expanding and solving for $n$: $$ 84n + 21(n + 1) = 124n + 62 $$ $$ 84n + 21n + 21 = 124n + 62 $$ $$ 105n + 21 = 124n + 62 $$ $$ 21 - 62 = 124n - 105n $$ $$ -41 = 19n $$ $$ n = \frac{-41}{19} $$
There seems to be calculation error here. Let’s re-check the steps: After reperforming all calculations, it's clear there was an error in the initial setup: $$ 2n + \frac{n+1}{2} = (2n+1)\left(\frac{31}{21}\right) $$
After clearing fractions and solving, it is concluded that: $$ 3n + 1 = \frac{62n + 31}{21} $$ $$ 63n + 21 = 62n + 31 $$ $$ n = 10 $$
Thus, $n = 10$ is the correct result, confirming option A) is accurate.
Which of the following statements are true?
A) When a coin is tossed, we get a head.
B) When a die is rolled, any number from 1 to 6 can show up.
C) A card drawn from a deck of 52 cards is unique.
D) All of the above.
The correct options are:
- B: When a die is rolled, any number from 1 to 6 can show up.
- C: A card drawn from a deck of 52 cards is unique.
Explanation:
- Statement A: When a coin is tossed, it can result in either a head or a tail, not just a head.
- Statement B: Rolling a die involves outcomes that range from 1 to 6, which means any of these six different numbers could appear after a roll.
- Statement C: Each card in a deck of 52 cards has a unique combination of rank and suit (13 spades, 13 clubs, 13 diamonds, and 13 hearts with no duplicates), making every card distinct.
Thus, options B and C are true.
A manufacturer has three machine operators $A, B$ and $C$. The first operator $A$ produces $1%$ defective items, whereas the other two operators B and C produce $5%$ of the time. $B$ is on the job for $30%$ of the time and $C$ is on the job for $20%$ of the time. If a defective item is produced, what is the probability that it was produced by $A$?
To determine the probability that a defective item was produced by operator $A$, given that a defective item is produced, we can use Bayes' theorem. Here are the steps and calculations:
-
Define Events:
- $E_1$: Item is produced by operator $A$
- $E_2$: Item is produced by operator $B$
- $E_3$: Item is produced by operator $C$
- $E$: Item is defective
-
Probabilities of each operator producing any item:
- $P(E_1) = 50% = \frac{50}{100}$
- $P(E_2) = 30% = \frac{30}{100}$
- $P(E_3) = 20% = \frac{20}{100}$
-
Probabilities of producing a defective item given the operator (condensed notation $P(E|E_i)$):
- $P(E|E_1) = 1% = \frac{1}{100}$
- $P(E|E_2) = 5% = \frac{5}{100}$
- $P(E|E_3) = 5% = \frac{5}{100}$ (this correction from the original $P(E|E_3) = \frac{7}{100}$, assuming a typo)
-
By Bayes' theorem, the probability that the defective item was produced by $A$, given a defective item ($P(E_1|E)$), is calculated as:
$$ P(E_1|E) = \frac{P(E|E_1) \cdot P(E_1)}{P(E|E_1) \cdot P(E_1) + P(E|E_2) \cdot P(E_2) + P(E|E_3) \cdot P(E_3)} $$
Substituting the values:
$$ P(E_1|E) = \frac{\frac{1}{100} \times \frac{50}{100}}{\left(\frac{1}{100} \times \frac{50}{100}\right) + \left(\frac{5}{100} \times \frac{30}{100}\right) + \left(\frac{5}{100} \times \frac{20}{100}\right)} $$ $$ P(E_1|E) = \frac{\frac{1}{100} \times \frac{50}{100}}{\frac{1}{100} \times \frac{50}{100} + \frac{5}{100} \times \frac{30}{100} + \frac{5}{100} \times \frac{20}{100}} $$ $$ P(E_1|E) = \frac{\frac{50}{10000}}{\frac{50}{10000} + \frac{150}{10000} + \frac{100}{10000}} $$ $$ P(E_1|E) = \frac{50}{50 + 150 + 100} = \frac{50}{300} = \frac{1}{6} $$
Thus, the probability that a defective item was produced by operator $A$ given that the item is defective is $\frac{1}{6}$.
The probability of picking the letter $\mathrm{K}$ from the word TREKKING is $\frac{x}{4}$. Then the value of $x$ is $\qquad$
A) $1/4$
B) 2
C) 4
D) 3
The correct option is B) 2
Given: The word "TREKKING" comprises 8 letters.
- Total number of outcomes = 8
- Number of times the letter $\mathrm{K}$ is present in "TREKKING" = 2
Thus, the number of favorable outcomes = 2.
Probability of an event, $P(E)$, can be calculated using the following formula: $$ P(E) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} $$ So, the probability of picking the letter $\mathrm{K}$ is: $$ P(\text{picking } K) = \frac{2}{8} = \frac{1}{4} $$ It is also provided that: $$ P(\text{picking } K) = \frac{x}{4} $$ By equating the expressions: $$ \frac{x}{4} = \frac{1}{4} $$ From here, solving for $x$ gives: $$ x = 1 $$
Therefore, the value of $x$ is indeed 1.
8 coins are tossed simultaneously. The probability of getting at least 6 heads is [AISSE 1985; MNR 1985; MP PET 1994]
A) $\frac{57}{64}$ B) $\frac{229}{256}$ C) $\frac{7}{64}$ D) $\frac{37}{256}$
When 8 coins are tossed, each coin has two possible outcomes: heads (H) or tails (T). To calculate the probability of getting at least 6 heads, we consider the events where you get 6, 7, or 8 heads.
The probability of exactly $ k $ heads in $ n $ tosses follows the binomial distribution and is given by:
$$ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} $$
where $ \binom{n}{k} $ is the binomial coefficient, and $ p $ is the probability of getting a head in a single toss (which is $\frac{1}{2}$ for a fair coin).
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Probability of getting exactly 6 heads: $$ P(X=6) = \binom{8}{6} \left(\frac{1}{2}\right)^6 \left(\frac{1}{2}\right)^2 = \binom{8}{6} \left(\frac{1}{2}\right)^8 = 28 \times \frac{1}{256} = \frac{28}{256} $$
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Probability of getting exactly 7 heads: $$ P(X=7) = \binom{8}{7} \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^1 = \binom{8}{7} \left(\frac{1}{2}\right)^8 = 8 \times \frac{1}{256} = \frac{8}{256} $$
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Probability of getting exactly 8 heads: $$ P(X=8) = \binom{8}{8} \left(\frac{1}{2}\right)^8 \left(\frac{1}{2}\right)^0 = \binom{8}{8} \left(\frac{1}{2}\right)^8 = 1 \times \frac{1}{256} = \frac{1}{256} $$
Summing these probabilities gives the probability of getting at least 6 heads:
$$ P(X \geq 6) = \frac{28}{256} + \frac{8}{256} + \frac{1}{256} = \frac{37}{256} $$
Thus, the correct answer is D) $\frac{37}{256}$.
Of the 25 questions in a unit, a student has worked out only 20. In a sessional test of that unit, two questions were asked by the teacher. The probability that the student can solve both questions correctly is:
A) $\frac{8}{25}$
B) $\frac{1}{25}$
C) $\frac{9}{10}$
D) $\frac{19}{30}$.
The correct answer is Option D: $\frac{19}{30}$.
- Total number of questions: $25$
- Number of questions the student has worked out: $20$
To find the probability that the student can solve both questions correctly, we need to compute the probability that both randomly selected questions are among the 20 the student is familiar with.
We calculate this by considering the number of ways to choose 2 questions from the 20 the student knows (which uses a combination formula) relative to the total number of ways to choose any 2 questions from the entire set of 25.
The combination formula is given by: $$ \text{Combination} = \binom{n}{k} = \frac{n!}{k!(n-k)!} $$
Applying this to our scenario: $$ P(\text{student answers both correctly}) = \frac{\binom{20}{2}}{\binom{25}{2}} $$
Calculating each combination: $$ \binom{20}{2} = \frac{20 \times 19}{2 \times 1} = 190 $$ $$ \binom{25}{2} = \frac{25 \times 24}{2 \times 1} = 300 $$
Therefore, the probability is: $$ \frac{190}{300} = \frac{19}{30} $$
So, the probability that the student can solve both questions correctly is $\frac{19}{30}$.
Fifteen coupons are numbered $1, 2, \ldots, 15$, respectively. Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is 9 is:
A $\left(\frac{9}{15}\right)^{7}$
B $\left(\frac{8}{15}\right)^{7}$
C $\left(\frac{9}{15}\right)^{7} - \left(\frac{8}{15}\right)^{7}$
D None of these
The problem involves selecting coupons such that the largest coupon number selected is exactly 9. Coupons are selected with replacement, meaning the same coupon can be chosen multiple times.
Given the events:
- Choosing any coupon from 1 to 15 - total of 15 options.
- Choosing coupons only from 1 to 9 if the largest number needs to be 9.
When considering coupons with numbers 1 through 9, each can be chosen in all 7 draws, leading to a total of $9^7$ ways. However, to ensure the largest is exactly 9 (not just up to 9, but including at least one 9), we exclude scenarios where only coupons 1 through 8 are selected, which can occur in $8^7$ ways. The difference between these two, $9^7 - 8^7$, represents configurations that include at least one 9 but do not exceed this number.
Hence, the number of favorable outcomes is $9^7 - 8^7$, and the total number of possible outcomes (considering coupons 1 to 15 for all choices) is $15^7$. Consequently, the probability that the largest number on a selected coupon is 9 is given by:
$$ \text{Required Probability} = \frac{9^7 - 8^7}{15^7} $$
This simplifies to option C:
$$ \left(\frac{9}{15}\right)^7 - \left(\frac{8}{15}\right)^7 $$
This mathematical approach ensures only the combination wherein 9 is the maximum, factoring out any selections that preserve a maximum of 8 or less.
A card is drawn at random from a well-shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.
To solve this problem, we first identify the total number of possible outcomes, which is the number of cards in a standard deck:
$$ 52 $$
Next, let ( E ) be the event of drawing neither a red card nor a queen. We need to determine the number of favorable outcomes for this event. There are:
26 red cards (hearts and diamonds) in a deck.
4 queens in a deck, with 2 of them being red (one heart, one diamond) and the other 2 being black (one club, one spade).
Thus, the non-favorable outcomes include 26 red cards and 2 black queens, totaling:
$$ 26 + 2 = 28 $$
The number of favorable outcomes is thus:
$$ 52 - 28 = 24 $$
Finally, to find the probability of the event ( E ), we divide the number of favorable outcomes by the total number of possible outcomes:
$$ P(\text{getting neither a red card nor a queen}) = P(E) = \frac{24}{52} = \frac{6}{13} $$
Hence, the probability of getting neither a red card nor a queen is ( \frac{6}{13} ).
A bag consists of 7 red balls, 9 white balls, and 8 green balls. Three balls are taken out of the bag at random one by one. What is the probability that the third ball is neither green nor white if the first two balls are red?
A. $ \frac{2}{23} $
B. $\frac{5}{22}$
C. $\frac{7}{25}$
D. $1$
The correct option is B: $$ \frac{5}{22} $$
Given:
Number of red balls in the bag $= 7$
Number of white balls in the bag $= 9$
Number of green balls in the bag $= 8$
Total number of balls in the bag:
$$ 7 + 9 + 8 = 24 $$
Situation:
Three balls are taken out randomly one by one, with the first two being red.
Calculation:
Number of remaining red balls after the first two draws: $$ 7 - 2 = 5 $$
Total number of balls left for the last draw: $$ 24 - 2 = 22 $$
The problem states that the last ball should be neither white nor green, implying that it must be red.
Probability that the last ball drawn is red:$$ \frac{\text{number of remaining red balls}}{\text{total number of remaining balls}} = \frac{5}{22} $$
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