Two different dice are thrown simultaneously.

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Two different dice are thrown simultaneously. Find the probability of getting:
(i) a number greater than 3 on each dice
(ii) an odd number on both dice.

Solution:

When two different dice are thrown simultaneously,
then the sample space S of the random experiment =
{(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)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) .
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
It consists of 36 equally likely outcomes.
(i) Let E be the event of ‘a number greater than 3 on each dice’.
E = {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
No. of favourable outcomes (E) = 9
P (number greater than 3 on each dice) = \\ \frac { 9 }{ 36 } = \\ \frac { 1 }{ 4 }
(ii) Let E be the event of ‘an odd number on both dice’.
E = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
No. of favourable outcomes (E) = 9
∴ P (Odd on both dices) = \\ \frac { 9 }{ 36 } = \\ \frac { 1 }{ 4 }

Two different dice are thrown at the same time. Find the probability of getting :
(i) a doublet
(ii) a sum of 8
(iii) sum divisble by 5
(iv) sum of atleast 11.

Solution:

Two different dice are thrown at the same time
Possible outcomes will be (6)² i.e. 36
(i) Number of events which doublet = 6
i.e. (1, 1), (2, 2) (3, 3), (4, 4), (5, 5) and (6, 6)
.’. Probability of doublets will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 6 }{ 36 }
= \\ \frac { 1 }{ 6 }
(ii) Number of event in which the sum is 8 are
(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) = 5
Probability of a sum of 8 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 5 }{ 36 }
(iii) Number of event when sum is divisible by
5 are (1, 4), (4, 1), (2, 3), (3, 2), (4, 6),
(5, 5) = 7 in numbers
Probability of sum divisible by 5 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 7 }{ 36 }
(iv) Sum of atleast 11, will be in following events
(5, 6), (6, 5), (6, 6)
Probability of sum of atleast 11 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 3 }{ 36 }
= \\ \frac { 1 }{ 12 }

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