Question

8 coins are tossed simultaneously. The probability of getting at least 6 heads is:

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Answer

To determine the probability of getting at least 6 heads when 8 coins are tossed simultaneously, we can use the binomial distribution formula. Here's a step-by-step solution to this problem:

Step 1: Understanding Probability with Binomial Distribution

  • The probability of getting a head in a single coin toss, $P$, is $\frac{1}{2}$.

  • Using the binomial distribution, we can calculate the probability of getting exactly $x$ heads out of $n$ tosses using the formula:

    $$ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} $$

    where $p$ is the probability of success (getting heads), $n$ is the number of trials (coin tosses), and $x$ is the number of successes (number of heads).

Step 2: Problem Requirement

We need the probability of getting at least 6 heads, which means we need the sum of the probabilities of getting 6, 7, or 8 heads. Hence,

$$ P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) $$

Step 3: Applying the Binomial Distribution Formula

For $P(X = 6)$:

$$ \binom{8}{6} \left(\frac{1}{2}\right)^6 \left(\frac{1}{2}\right)^{8-6} = \binom{8}{6} \left(\frac{1}{2}\right)^8 $$

For $P(X = 7)$:

$$ \binom{8}{7} \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^{8-7} = \binom{8}{7} \left(\frac{1}{2}\right)^8 $$

For $P(X = 8)$:

$$ \binom{8}{8} \left(\frac{1}{2}\right)^8 \left(\frac{1}{2}\right)^{8-8} = \binom{8}{8} \left(\frac{1}{2}\right)^8 $$

Step 4: Calculating the Values

  • $\binom{8}{6} = 28$
  • $\binom{8}{7} = 8$
  • $\binom{8}{8} = 1$

Using these values in the formula, we get:

$$ P(X \geq 6) = \left( 28 \times \left(\frac{1}{2}\right)^8 \right) + \left( 8 \times \left(\frac{1}{2}\right)^8 \right) + \left( 1 \times \left(\frac{1}{2}\right)^8 \right) $$

Step 5: Factoring and Summing Up

Factoring $\left(\frac{1}{2}\right)^8$ out:

$$ P(X \geq 6) = \left(\frac{1}{2}\right)^8 \left( 28 + 8 + 1 \right) = \left(\frac{1}{256}\right) \times 37 = \frac{37}{256} $$

Final Answer

The probability of getting at least 6 heads when 8 coins are tossed simultaneously is:

$$ \boxed{\frac{37}{256}} $$


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