IIT JEE , AIEEE Forum - Mathematics , Algebra

trisha

A bag contains n+1 coins .It is known that one of these coins shows head on both sides,whereas other coins are fair.one coin is selected at random and tossed.If the probability that the toss results in head is 7/12.Find the value of n.

rik_mad

hi,
probability of choosing the coin with both heads is
1/(n+1)
probability that it gives head is
[1/(n+1)](1) ......(1)
probability of choosing the rest of the coins is
[n/n+1]
probability that the chosen coins give head is
[n/n+1](1/2) ......(2)

adding (1) and (2) and equating to 7/12
we get n as 5
and n+1 as 6.

trisha

THANKS

iitkgp_bipin

Probability of choosing the biased coin = 1/(n+1)
Probability of coming head on tossing biased coin = 1

Probability of choosing fair coin = n/(n+1)
Probability of coming head on tossing fair coin = 1/2

Hence probability of coming head = 1/(n+1) + [n/(n+1)].[1/2] = (n+2)/(2n+2)

Given this is equal to 7/12.
Hence (n+2)/(2n+2) = 7/12

Gives n = 5.

Best Wishes

trisha

thank uall

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