The chance that a student will clear the quiz-11 paper is 0.60.6 and the chance that he will clear both quiz-11 and quiz-22 papers is 0.270.27. The chance that he will clear at least one quiz paper is 0.830.83. What is the chance that he will clear quiz-22 paper?(Enter the answer correct to 2 decimal accuracy)
Question
The chance that a student will clear the quiz-11 paper is 0.60.6 and the chance that he will clear both quiz-11 and quiz-22 papers is 0.270.27. The chance that he will clear at least one quiz paper is 0.830.83. What is the chance that he will clear quiz-22 paper?(Enter the answer correct to 2 decimal accuracy)
Solution
To solve this problem, we can use the formula for the probability of the union of two events, which is P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Here, A is the event of clearing quiz-1, B is the event of clearing quiz-2, and A ∩ B is the event of clearing both quizzes.
Given in the problem, we have:
P(A) = P(quiz-1) = 0.60 P(A ∩ B) = P(quiz-1 and quiz-2) = 0.27 P(A ∪ B) = P(at least one quiz) = 0.83
We can substitute these values into the formula to find P(B), the probability of clearing quiz-2:
0.83 = 0.60 + P(B) - 0.27 P(B) = 0.83 - 0.60 + 0.27 P(B) = 0.50
So, the chance that the student will clear the quiz-2 paper is 0.50 or 50%.
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