Q18 Comprehension 2 Marks 3 Sep 2023
Passage
A six-sided die is marked with ' 1 1 1 ' on two of its faces, ' 2 2 2 ' on one of its faces, and ' 3 3 3 ' on the remaining three faces. The die is thrown twice independently. Let X X X denote the total sum of scores in the two throws. Based on the given information, answer the subquestions.
What is the value of P ( X > 2 ∣ X < 5 ) P(X > 2 \mid X < 5) P ( X > 2 ∣ X < 5 ) ? A B C D 23 36 F r o m t h e P M F o f X c a l c u l a t e d i n t h e p r e v i o u s p a r t : − P ( X = 2 ) = 1 9 = 4 36 − P ( X = 3 ) = 1 9 = 4 36 − P ( X = 4 ) = 13 36 S o , P ( X < 5 ) = P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 ) = 4 36 + 4 36 + 13 36 = 21 / 36. A n d , P ( 2 < X < 5 ) = P ( X = 3 ) + P ( X = 4 ) = 4 36 + 13 36 = 17 / 36. T h e r e f o r e , P ( X > 2 ∣ X < 5 ) = ( 17 / 36 ) / ( 21 / 36 ) = 17 / 21. \frac{23}{36} From the PMF of X calculated in the previous part: - P(X=2) = \frac{1}{9} = \frac{4}{36} - P(X=3) = \frac{1}{9} = \frac{4}{36} - P(X=4) = \frac{13}{36} So, P(X < 5) = P(X=2) + P(X=3) + P(X=4) = \frac{4}{36} + \frac{4}{36} + \frac{13}{36} = 21/36. And, P(2 < X < 5) = P(X=3) + P(X=4) = \frac{4}{36} + \frac{13}{36} = 17/36. Therefore, P(X > 2 \mid X < 5) = (17/36) / (21/36) = 17/21. 36 23 F r o m t h e P M F o f X c a l c u l a t e d in t h e p r e v i o u s p a r t : − P ( X = 2 ) = 9 1 = 36 4 − P ( X = 3 ) = 9 1 = 36 4 − P ( X = 4 ) = 36 13 S o , P ( X < 5 ) = P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 ) = 36 4 + 36 4 + 36 13 = 21/36. A n d , P ( 2 < X < 5 ) = P ( X = 3 ) + P ( X = 4 ) = 36 4 + 36 13 = 17/36. T h er e f or e , P ( X > 2 ∣ X < 5 ) = ( 17/36 ) / ( 21/36 ) = 17/21. 