Question

29 multiple choice 2 points if a fair coin is tossed five times, the probability of getting exactly two heads is 2/5 10/32 1/32 1/4

Explanation:

Step1: Calculate the number of possible outcomes

When a coin is tossed \(n\) times, the total number of possible outcomes is \(2^n\). Here \(n = 5\), so the total number of possible outcomes is \(2^5=32\).

Step2: Calculate the number of ways to get exactly 2 heads

The number of ways to choose \(k\) successes (in this case, getting heads) from \(n\) trials is given by the binomial coefficient \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\). Here \(n = 5\) and \(k=2\), so \(\binom{5}{2}=\frac{5!}{2!(5 - 2)!}=\frac{5\times4\times3!}{2\times1\times3!}=10\).

Step3: Calculate the probability

The probability \(P(X = k)\) of getting \(k\) successes in \(n\) independent Bernoulli trials (coin - tosses in this case) is \(P(X = k)=\binom{n}{k}p^{k}(1 - p)^{n - k}\). Since \(p=\frac{1}{2}\) (probability of getting a head in a single fair - coin toss), \(P(X = 2)=\binom{5}{2}(\frac{1}{2})^{2}(1-\frac{1}{2})^{5 - 2}\). Substituting \(\binom{5}{2}=10\), \(p=\frac{1}{2}\), we get \(P(X = 2)=10\times(\frac{1}{2})^{2}\times(\frac{1}{2})^{3}=10\times\frac{1}{32}=\frac{10}{32}\).

Answer:

\(\frac{10}{32}\) (corresponding to the second option)