Suppose that a family has 8 children.​ Also, suppose that the probability of having a boy is one half 1/2 . Find the probability that the family has no more than 6 boys.

Question

answer in simplified fraction form and use the binomial probability formula

Answer 1

Probability Of 6 Male Children At Most is written as P(X≤6).

P(X≤6) equals ₈C₀(1/2)⁰(1/2)⁸ + ₈C₁(1/2)¹(1/2)⁷ + ₈C₂(1/2)²(1/2)⁶ + ₈C₃(1/2)³(1/2)⁵ + ₈C₄(1/2)⁴(1/2)⁴ + ₈C₅(1/2)⁵(1/2)³ + ₈C₆(1/2)⁶(1/2)² where _nC_s(p)^s(1-p)^(n-s) translates to n!÷s!÷(n-s)!×p^s×(1-p)^(n-s).

The string of 7 combinations above simplifies to (1/2)⁸ times [₈C₀ + ₈C₁ + ₈C₂ + ₈C₃ + ₈C₄ + ₈C₅ + ₈C₆] or (1/256)[1+8+28+56+70+56+28] or 247/256.

P(X≤6) = 247/256 is also given by 1 − [₈C₇(1/2)⁷(1/2)¹ + ₈C₈(1/2)⁸(1/2)⁰] or 1 − (1/2)⁸[8 + 1] or 1 − 9/256 equal to 247/256.