A fair coin is tossed 3 times. Find the probability of obtaining exactly 2 heads.
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Correct Answer: Option B
Explanation:
Let the probability of getting a head = p = \(\frac{1}{2}\) and that of tail = q = \(\frac{1}{2}\)
\((p + q)^{3} = p^{3} + 3p^{2}q + 3pq^{2} + q^{3}\)
In the equation above, \(p^{3}\) and \(q^{3}\) are the probabilities of 3 heads and 3 tails respectively
while, \(p^{2}q\) and \(pq^{2}\) are the probabilities of 2 heads and one tail and 2 tails and one head respectively.
Probability of exactly 2 heads = \(3p^{2}q = 3(\frac{1}{2})^{2}(\frac{1}{2})\)
= \(\frac{3}{8}\)
Let the probability of getting a head = p = \(\frac{1}{2}\) and that of tail = q = \(\frac{1}{2}\)
\((p + q)^{3} = p^{3} + 3p^{2}q + 3pq^{2} + q^{3}\)
In the equation above, \(p^{3}\) and \(q^{3}\) are the probabilities of 3 heads and 3 tails respectively
while, \(p^{2}q\) and \(pq^{2}\) are the probabilities of 2 heads and one tail and 2 tails and one head respectively.
Probability of exactly 2 heads = \(3p^{2}q = 3(\frac{1}{2})^{2}(\frac{1}{2})\)
= \(\frac{3}{8}\)