Probability - 3

Bagrang's relation ship





Dumb Question :- How relation P (getting r' s success)

= P^r q^n-r nC _r

Ans - In this P is probability of getting success and q is probability of getting failure and p + q = 1

Sclection of r success out of n out comes = nC_r

img..... P(getting r' s success)

= P x P x P...............r times x q x q x.............(n - r) times x nC_r

Illustration :- A die is rolled n times. Find the value of n if probability of getting at least one ace is 91/216

Ans P(getting at least one ace) = 1 - P(no ace)

P(1) = 1/6 P(not 1) = 5/6



=> n = 3

Probability theorem - (Baye's theorem)

If an event A can occur with n mutually exclusive, exhaustive
event B₁, B₂, ..............B_n and probability of P(A/B₁),
P(A/B₂), ................P(A/B_n) is known. Then probability .

P(Bi/A) =

P(A img B_i) = P(Bi). P(A/Bi)
= P (A). P(Bi/A)



P(A) = P(A img B₁) + P(A img B₂) + ......+ P(B_n) P(A/B_n)

= P(B_i) P(A / B_i)

=P(A). P(B_i / A)

P(A) =

=P(B₁) P(A / B₁) P (A / B₂) + ............+ P(B_n) P(A / B_n)

P(B_r) P (A / B_r)

Illustration :- A bag cintains 5 balls of unknown color. Two balls are drqwn and both are found to be white. Find the chance that there are 4 white ball in box . Assume any no. of white ball equally likely .

Ans - A : Two balls drawn from a bag contuins 5 balls.
both are white
B₀ : 0 white ball
B₁ : 1 white ball
B₂ : 2 white ball
B₃ : 3 white ball
B₄ : 4 white ball
B₅ : 5 white ball
P(B₀) = P(B₀) = P(B₁) = P(B₃) = P(B₄) =P(B₅) =P(B₆) = 1/6

P(A/B₀) = 0 P(A/B₁) = 0

P(A/B₂) =

P(A/B₃) =

P(A/B₄) =

P(A/B₅) =

P(B₄/A) =

=