study material-mathematics-algebra-probability

# 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 B

_{1}, B

_{2}, ..............B

_{n}and probability of P(A/B

_{1}),

P(A/B

_{2}), ................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 B

_{1}) + P(A img B

_{2}) + ......+ P(B

_{n}) P(A/B

_{n})

= P(B

_{i}) P(A / B

_{i})

=P(A). P(B

_{i}/ A)

P(A) =

=P(B

_{1}) P(A / B

_{1}) P (A / B

_{2}) + ............+ 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

B

_{0}: 0 white ball

B

_{1}: 1 white ball

B

_{2}: 2 white ball

B

_{3}: 3 white ball

B

_{4}: 4 white ball

B

_{5}: 5 white ball

P(B

_{0}) = P(B

_{0}) = P(B

_{1}) = P(B

_{3}) = P(B

_{4}) =P(B

_{5}) =P(B

_{6}) = 1/6

P(A/B

_{0}) = 0 P(A/B

_{1}) = 0

P(A/B

_{2}) =

P(A/B

_{3}) =

P(A/B

_{4}) =

P(A/B

_{5}) =

P(B

_{4}/A) =

=