Question

A black and a red die are rolled.

1. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
2. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Answer 1

$(a)$ n(S) = 6 x 6 = 36

Let, A represents obtaining a sum greater than 9 and B represents black die resulted in a 5.

A = (46, 64, 55, 36, 63, 45, 54, 65, 56, 66)

⇒n(A)=10

$P(A)=\frac {n(A)}{n(S)}=\frac {10}{36}$

B=(51, 52, 53, 54, 55, 56)

⇒n(B) = 6

$P(B)=\frac {n(B)}{n(S)} =\frac {6}{36}$

A∩B = (55, 56)

⇒$n(A∩B)=\frac {2}{36}$

$P(A|B)=\frac {P(A∩B)}{P(B)}$

$P(A|B)=\frac {2/36}{6/36}$

$P(A|B)=\frac 26$

$P(A|B)=\frac 13$

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$(b)$ Let, A denote the sum as 8

∴ A = {(2, 6), (3, 5), 4, 4), (5, 3), (6, 2)}

B = Red die results in a number less than 4, either first or second die is red

∴B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}

$P(B)=\frac {n(B)}{n(S)}$

$P(B)=\frac {18}{36}$

$P(B)=\frac 12$

A∩B={(2, 6),(3, 5)}

⇒$n(A∩B)=2$

$P(A∩B) =\frac {2}{36}=\frac {1}{18}$

$P(A|B)=\frac {P(A∩B)}{P(B)}$

$P(A|B)=\frac {1/18}{1/2}$

$P(A|B)=\frac {2}{18}$

$P(A|B)=\frac 19$

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