How do you find the probability of obtaining at least one tail when a coin is tossed five times?

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How do you find the probability of obtaining at least one tail when a coin is tossed five times?

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Answer 1 · 2017-05-01T11:13:39

$1 - \frac{1}{32} = \frac{31}{32}$ $1 - 1/32 = 31/32$

Explanation:

'At least one tail' means that there can be one, or two or three or four or five tails.

The only option that is not included is five heads.

The sum of all the probabilities is always $1$ $1$ .

$P (at least one tail) = 1 - P (no tail)$ $P("at least one tail") = 1 - P("no tail")$

$P (no tail) = P (all heads) = P (H,H,H,H,H)$ $P("no tail") = P("all heads") = P("H,H,H,H,H")$

In any throw the probability of a tail or a head is $\frac{1}{2}$ $1/2$

$P (H,H,H,H,H) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = {(\frac{1}{2})}^{5} = \frac{1}{32}$ $P("H,H,H,H,H") = 1/2xx1/2xx1/2xx1/2xx1/2 = (1/2)^5 = 1/32$

$P (at least one tail) = 1 - \frac{1}{32} = \frac{31}{32}$ $P("at least one tail") = 1 - 1/32 = 31/32$

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How do you find the probability of obtaining at least one tail when a coin is tossed five times?

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