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Read the passage and answer the questions.
Rahul and Divya were playing the snakes and ladders board game. Each one had their own dice to play…
Rahul and Divya were playing the snakes and ladders board game. Each one had their own dice to play the game. Rahul was using a red dice, whereas Divya was using a black dice. In the beginning of the game, they were using their own dice to play. After some time, in order to play the game faster they both started using both the dice together for playing.
When Divya rolled both red and black dice together then:
(a)[2.0]
find the conditional probability of obtaining sum greater than 9, given that black dice resulted in a 5.
(b)[2.0]
find the conditional probability that sum of the number on the dice is not 4, given that the numbers on the both the dice are different.
Answer
Answer (a)
Official answer keyLet $A$ represent obtaining sum greater than 9 and $B$ represents black dice resulted in a 5.
$n(S) = 36$
$n(A) = \{(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)\} = 6$
$n(B) = \{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\} = 6$
$n(A \cap B) = \{(5,5), (5,6)\} = 2$
$P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{2/36}{6/36} = \frac{1}{3}$.
Final answer: $\frac{1}{3}$
Answer (b)
Official answer keyLet $A$ represent obtaining sum is 4 and $B$ represents both the dice show different number.
$n(S) = 36$, $n(A) = \{(1,3), (2,2), (3,1)\} = 3$, $n(B) = 30$, $n(A\cap B) = \{(1,3), (3,1)\} = 2$
$P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{2/36}{30/36} = \frac{1}{15}$.
P(sum of the numbers is not 4 | different numbers) $P(A'|B) = 1 - P(A|B) = 1 - \frac{1}{15} = \frac{14}{15}$
Final answer: $\frac{14}{15}$
From ISC 2025 Practice Mathematics, question 109.