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If a certain coin is flipped, the probability that the coin will land heads up is \frac{1}{2} . If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

Explanation	Calculations	Help
$We first translate "the probability that the coin will land heads up is 1/2", giving the probability of heads and tails.$	$P (H) = \frac{1}{2} P (T) = 1 - \frac{1}{2} = \frac{1}{2}$	Theory & Tactics Method Card TRANS1-B Click to view full details
$We then multiply the probabilities of independent flips to find the probability that the coin lands heads on the first three flips and tails on the last two flips.$	$P (H then H then H then T then T) = P (H)^{3} \times P (T)^{2} = (\frac{1}{2})^{3} \times (\frac{1}{2})^{2} = (\frac{1}{2})^{5} = \frac{1}{32}$	Theory & Tactics Method Card PC1-C Click to view full details

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\frac{1}{32}

If a certain coin is flipped, the probability that the coin will land heads up is \frac{1}{2} . If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

Explanation	Calculations	Help
$We first translate "the probability that the coin will land heads up is 1/2", giving the probability of heads and tails.$	$P (H) = \frac{1}{2} P (T) = 1 - \frac{1}{2} = \frac{1}{2}$	Theory & Tactics Method Card TRANS1-B Click to view full details
$We then multiply the probabilities of independent flips to find the probability that the coin lands heads on the first three flips and tails on the last two flips.$	$P (H then H then H then T then T) = P (H)^{3} \times P (T)^{2} = (\frac{1}{2})^{3} \times (\frac{1}{2})^{2} = (\frac{1}{2})^{5} = \frac{1}{32}$	Theory & Tactics Method Card PC1-C Click to view full details

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\frac{1}{32}