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Bunuel
When a fair coin is tossed 6 times, what is the probability that less than 5 heads are tossed?

A. 7/64
B. 5/16
C. 2/3
D. 49/64
E. 57/64


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5H 1T( !6/!5*!1) or 6H(!6/!6)
Total cases= 6+1=7
P(not required)=7/64
P(required)=1-7/64
=57/64
E:)
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P of atleast 5tails = 6c6*(1/2)^6 + 6c5*(1/2)^6 ; (1/64+6/64) ; 7/64
P of Heads<5 ; 1-7/64 ; 57/64
IMO E


Bunuel
When a fair coin is tossed 6 times, what is the probability that less than 5 heads are tossed?

A. 7/64
B. 5/16
C. 2/3
D. 49/64
E. 57/64


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We need to find When a fair coin is tossed 6 times, what is the probability that less than 5 heads are tossed?

Coin is tossed 6 times => Total number of cases = \(2^6\) = 64

P(Less than 5 heads) = 1 - (P(5H) - P(6H)

P(5H)

Cases in which we get 5H is same as getting 1 Tail and this can be found by putting 1 tail in any of the 6 slots _ _ _ _ _ _ _
This can be done in 6 ways by putting T in any of the 6 slots => 6 ways

=> P(5H) = \(\frac{6}{64}\)

P(6H)

This can be done in 1 way (HHHHHH)

=> P(6H) = \(\frac{1}{64}\)

P(Less than 5 heads) = 1 - (P(5H) - P(6H) = 1 - \(\frac{6}{64}\) - \(\frac{1}{64}\) = \(\frac{64 - 7}{64}\) = \(\frac{57}{64}\)

So, Answer will be E
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

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