A probability involving lots of children.

Probability Level 3

The probability that a family has exactly n n children is α p n , n ≥ 1 \alpha p^n, n \ge 1 and α \alpha is a positive constant. All sex distributions (male or female) of n n children in a family have the same probability. If the probability that a family contains exactly k k boys is

P k = A α B p k ( C − D p ) k + E P_k = \dfrac{A \alpha^B p^k}{ {\left( C - Dp \right)}^{k + E}}

where uppercase alphabets from A A to E E are positive integers. Find the value of A + B + C + D + E A+B+C+D+E


The answer is 7.

Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

The required probability is :

∑ n = 0 ∞ \displaystyle \sum_{n=0}^\infty (Probability that there are exactly n + k n+k children) × \times (Probability that exactly k k children out of n + k n+k children are boys)

= ∑ n = 0 ∞ α ( n + k n ) ( p 2 ) n + k =\displaystyle \sum_{n=0}^\infty \alpha \dbinom{n+k}{n}\left(\dfrac{p}{2}\right)^{n+k}

= α ( p 2 ) k ( 1 − p 2 ) − ( k + 1 ) =\alpha \left(\dfrac{p}{2}\right)^{k}\left(1-\dfrac{p}{2}\right)^{-(k+1)}

= 2 α p k ( 2 − p ) k + 1 =\boxed{\dfrac{2\alpha p^k}{\left(2-p\right)^{k+1}}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

@Tapas Mazumdar what is your average rank in AITS ?

A Former Brilliant Member - 3 years, 3 months ago

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I guess it would be near 1000 from the left side. Because in one of the AiTS, I got as bad as 2500 and this time I was under 300. The modal rank is in between 400-600.

Tapas Mazumdar - 3 years, 3 months ago

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