Question

In a single throw of a pair of different dice, what is the probability of getting (i) a prime no. on each dice? (ii)a total of 9 or 11?

Answer 1

As we throw two die we have 6^2 i.e 36 chances
So n(S) = 36
Let the event A be having prime on each dice
So chances are {(2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5)}
Hence n(A) = 9
Thus P(A) = 9/36 = 1/4
Let B be the event having the sum of 9 or 11
So chances will be like {(3,6), (6.3), (4.5), (5,4), (5,6), (6,5)}
So n(B) = 6
Thus P(B) = 6/36 = 1/6

Answer 2

(i) Favourable outcomes are (2, 2) (2, 3) (2, 5) (3, 2)   (3, 3) (3, 5) (5, 2) (5, 3) (5, 5) = 9 outcomes

P (a prime number on each die) = $\bf\huge\frac{9}{36}$

=  $\bf\huge\frac{1}{4}$

(ii) Favourable outcomes are (3, 6) (4, 5) (5, 4)  (6, 3) (5, 6) (6, 5) i.e. 6 outcomes

P (a total of 9 or 11) = $\bf\huge\frac{6}{36}$

= $\bf\huge\frac{1}{6}$