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Question

Two persons A and B have respectively n+1 and n coins, which they toss simultaneously. Then the probability that A will have more heads than B is:


A

12

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B

>12

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C

<12

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D

None of these

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Solution

The correct option is A

12


Explanation for the correct option:

Let a and a' be the number of heads and tails by A.

So, a+a'=n+1 ...(1)

Let b and b' be the number of heads and tails by B.

So, b+b'=n

Required probability is that A has more heads than B, so let, P(a>b)=P(a'>b')=p

⇒P(a>b)=1-P(a≤b)

Now for, a≤b

⇒n+1-a'≤n-b'⇒1+b'≤a'

Thus, b'<a', so we can say that P(a≤b)=P(b'<a')

⇒P(a>b)=1-P(b'<a')⇒p=1-p⇒2p=1⇒p=12

Therefore, option (A) is the correct answer.


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