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Re: A fair coin is tossed 5 times. What is the probability that [#permalink]

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06 Aug 2014, 20:43

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This post received KUDOS

Probability of all tails = 1/32 Probability of 4 tails = 5/32 ------------------------------------------------------------------------------ Probability of all tails or 4 tails = 1/32 + 5/32 = 6/32 = (3/16) ------------------------------------------------------------------------------

Therefore probability of at least 2 heads = 1- Prob of all or 4 tails = 1- (3/16) = 13/16

Re: A fair coin is tossed 5 times. What is the probability that [#permalink]

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07 Aug 2014, 10:22

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Fair coin is tossed 5 times . Hence total number of outcomes = 2^5 = 32.

Problem asks for probability of getting atleast heads twice. Hence if we calculate probability of getting Heads exactly once and probability of not getting Heads at all and subract it from the total probability of the event which is 1 (As total probability of certain event will be always 1) we can get the probability of getting atleast heads twice.

Probability of getting exactly one head and no heads= (Number of possible outcomes [ HTTTT , THTTT, TTHTT, TTTHT, TTTTH , TTTTT] = 6)/ (Total possible outcomes = 32)

=> 6/32 = 3/16

Hence probability of getting atleast heads twice = 1 - (3/16) = 13/16 => Choice [D]

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