Question

A student appeared for two quizzer. He is told that his chance of winning quiz-1 is 0.5, losing quiz-2 is 0.3 and losing both the quizzers is 0.2. Find the probability that the student will win quiz-2 when he has already won quiz-1.

Answer 1

Answer:

0.8

Step-by-step explanation:

Concept:

1.Addition theorem of probability

P(A∪B) = P(A) + P(B) - P(A∩B)

2. P(A') = 1 - P(A)

3. P(A/B) = P(A∩B)/P(B)

Let A be the event of winning quiz-1.

Let B be the event of winning quiz-2

Given:

P(A) = 0.5 , P(B') = 0.3 & P(A'∩B')= 0.2

P(B') = 0.3

1 - P(B) = 0.3

P(B) = 1 - 0.3

P(B) = 0.7

P(A'∩B')= 0.2

P[(A∪B)']= 0.2

1 - P(A∪B) =0.2

P(A∪B) = 1 - 0.2

P(A∪B) =0.8

By addition theorem of probability

P(A∪B) = P(A) + P(B) - P(A∩B)

0.8 = 0.5 + 0.7 - P(A∩B)

P(A∩B) = 1.2 - 0.8

P(A∩B) = 0.4

P(the student will win quiz-2 when he has already won quiz-1)

=P(B/A)

=P(A∩B)/P(A)

= 0.4/0.5

= 4/5

=0.8