Question

In a single throw of a pair of different dice. Find the probability of getting A prime number on each dice 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

P(of getting a prime no on each dice) = 1 / 4

P(of getting a total of 9 or 11) = 1 / 6

Step-by-step explanation:

Total no of possible outcomes in two dice = 6 * 6 = 36

Sample Space:

Possible outcomes for a prime no = (2,2), (2, 3), (2, 5), (3, 2), (3, 3),(3, 5), (5, 2), (5, 3), (5,5)

There are total 9 possible outcomes.

$\fbox{P(E) = No Of Outcomes / Total No Of Outcomes}$

(i) P(E) = 9 / 36

= 3 / 12

= 1 / 4

Favorable outcomes when sum of numbers are 9 or 11 are (3, 6), (4, 5), (5, 4), (5, 6),(6, 3), (6,5),

There are total 6 possible outcomes.

(ii) P(E) = 6 / 36

= 1 / 6