Question

Given that the two numbers appearing on throwing two dice are different.Find the probability of the event ‘the sum of numbers on the dice is 4’.

Answer 1

When dice is thrown, number of observations in the sample space = 6 × 6 = 36

S = (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)
(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)
(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)
(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)
(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)
(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)

∴n(S)= 36

Let,
A represents the event “the sum of numbers on the dice is 4” and,
B represents the event “the two numbers appearing on throwing two dice are different”.

Therefore,
A = {(1, 3), (2, 2), (3, 1)}
⇒n(A)=3

$P(A)=\frac {n(A)}{n(S)}$
$P(A)=\frac {3}{36}$

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

$⇒ n(B)=30$

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

$P(B) =\frac {30}{36}$

Now,
A∩B = {(1,3),(3,1)}
⇒n(A∩C) = 2

$P(A∩B) =\frac {n(A∩B)}{n(S)}$

$P(A∩B)=\frac {2}{36}$

Hence, $P(A|B)=\frac {P(A∩B)}{P(B)}$

$P(A|B)=\frac {2/36}{30/36}$

$P(A|B)=\frac {2}{30}$

$P(A|B)=\frac {1}{15}$

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