Question -
Answer -
Let us assume probability of a randomly chosen studentpassing the first examination is 0.8 be P(A).
And also assume the probability of passing the secondexamination is 0.7 be P(B)
Then,
P(A∪B) is probability of passing at least one of theexamination
Now,
P(A∪B) = 0.95 , P(A)=0.8, P(B)=0.7
∴ P(A∪B) = P(A) + P(B) – P(A∩B)
0.95 = 0.8 + 0.7 – P(A∩B)
Transposing – P(A ∩ B) to LHS and it becomes P(A ∩ B) and0.95 to RHS and it becomes
– 0.95
P(A∩B) = 1.5 – 0.95
= 0.55
Hence, 0.55 is the probability that student will pass boththe examinations.