Two dice are rolled. If both dice have six fa | vedclass

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(A) The sample space $S$ consists of all possible pairs $(x, y)$ where $x, y \in \{1, 2, 3, 5, 7, 11\}$.Since there are $6$ faces on each die,the tota

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$\frac{1}{4}$

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(A) The sample space $S$ consists of all possible pairs $(x, y)$ where $x, y \in \{1, 2, 3, 5, 7, 11\}$.Since there are $6$ faces on each die,the total number of outcomes is $n(S) = 6 	imes 6 = 36$.Let $A$ be the event that the sum of the numbers is a prime number.The possible sums are:$1+1=2$ (prime),$1+2=3$ (prime),$1+3=4$,$1+5=6$,$1+7=8$,$1+11=12$$2+1=3$ (prime),$2+2=4$,$2+3=5$ (prime),$2+5=7$ (prime),$2+7=9$,$2+11=13$ (prime)$3+1=4$,$3+2=5$ (prime),$3+3=6$,$3+5=8$,$3+7=10$,$3+11=14$$5+1=6$,$5+2=7$ (prime),$5+3=8$,$5+5=10$,$5+7=12$,$5+11=16$$7+1=8$,$7+2=9$,$7+3=10$,$7+5=12$,$7+7=14$,$7+11=18$$11+1=12$,$11+2=13$ (prime),$11+3=14$,$11+5=16$,$11+7=18$,$11+11=22$The favorable outcomes are: $(1,1), (1,2), (2,1), (2,3), (2,5), (2,11), (3,2), (5,2), (11,2)$.Thus,$n(A) = 9$.The probability $P(A) = \frac{n(A)}{n(S)} = \frac{9}{36} = \frac{1}{4}$.

Two dice are rolled. If both dice have six faces numbered $1, 2, 3, 5, 7, 11$,then the probability that the sum of the numbers on the uppermost faces is a prime number is:

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