Question

For an urn containing 4 red balls and 6 green balls, let the number of balls randomly drawn be the number of heads turning up when 5 fair coins have been previously flipped. What is the probability of drawing 3 green balls?

why do we need binomial theorem for this

Answer 1

(we require binomial distribution for number of heads which on 5 flips since number of trails n=5 are fixed and probability of head is constant and independent from trail to trail)

probability of drawing 3 green balls

=P(3 heads)*P(3 green|3 heads)+P(4 heads)*P(3 green|4 heads)+P(5 heads)*P(3 green|5 heads)

=(₅C₃)*(0.5)³*(0.2)²*((₆C₃)*(₄C₀)/(₁₀C₃))+(₅C₄)*(0.5)⁴*(0.2)¹*((₆C₃)*(₄C₁)/(₁₀C₄))+(₅C₅)*(0.5)⁵*(0.2)⁰*((₆C₃)*(₄C₂)/(₁₀C₅))

=0.3125*0.1667+0.15625*0.3810+0.03125*0.4762

=0.1265