Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
2.7k views
in Probability by (28.2k points)
closed by

A black and a red dice are rolled

  1. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
  2. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

1 Answer

+1 vote
by (28.9k points)
selected by
 
Best answer

We have, n(S) = 36

1. E = Event of 5 on black die.

E = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

P(E) = \(\frac{6}{36} = \frac{1}{6}\)

F = Getting a sum greater than 9.

F = {(4, 6), (5, 5), (6, 4)(5, 6), (6, 5), (6, 6)}

⇒ E ∩ F = {(5,5), (5,6)}

P(E ∩ F) = \(\frac{2}{36} = \frac{1}{18}\)

Therefore the required probability

P(E/F) = \(\frac{P(E ∩ F)}{P(F)}\)\(\frac{\frac{1}{18}}{\frac{1}{6}} = \frac{1}{3}\)

2. E = Event of a number less than 4 on red die.

E = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3),

(3, 1), (3, 2), (3, 3), (3, 1), (3, 2), (3, 3),

(5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}

P(E) = \(\frac{18}{36}= \frac{1}{2}\)

F = Getting a sum 8.

F = {(4, 4), (5, 3), (3, 5)(2, 6), (6, 2), (6, 6)}

⇒ E ∩ F = {(5, 3),(6, 2)}

P(E ∩ F) = \(\frac{2}{36} = \frac{1}{18}\)

Therefore the required probability

 P(E/F) = \(\frac{P(E ∩ F)}{P(F)}\)\(\frac{\frac{1}{18}}{\frac{1}{2}} = \frac{1}{9}\).

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...