The probability that a student passes Biology is 0.35 and the probability to pass Chemistry is 0.63. If the probability to pass both Biology and Chemistry is 0.4, what is the probability that he will pass either Biology or Chemistry?

Prob(B or C) = Prob(B) + Prob(C) - Prob(B and C)

= .35+.63 - .4
= ...

0.58

To find the probability that the student will pass either Biology or Chemistry, we can use the principle of inclusion-exclusion.

Let's denote the probability of passing Biology as P(B) = 0.35 and the probability of passing Chemistry as P(C) = 0.63.

The probability of passing both Biology and Chemistry, P(B ∩ C), is given as 0.4.

The probability of passing either Biology or Chemistry can be calculated as follows:

P(B ∪ C) = P(B) + P(C) - P(B ∩ C)

P(B ∪ C) = 0.35 + 0.63 - 0.4

P(B ∪ C) = 0.98 - 0.4

P(B ∪ C) = 0.58

Therefore, the probability that the student will pass either Biology or Chemistry is 0.58 or 58%.

To find the probability that a student will pass either Biology or Chemistry, we can use the principle of addition. The formula for the probability of the union of two events is:

P(A or B) = P(A) + P(B) - P(A and B)

In this case, we are given the probabilities of passing Biology (P(Bio)) and passing Chemistry (P(Chem)). We are also given the probability of passing both Biology and Chemistry (P(Bio and Chem)).

Let's substitute the values we know into the formula:

P(Bio or Chem) = P(Bio) + P(Chem) - P(Bio and Chem)

P(Bio or Chem) = 0.35 + 0.63 - 0.4

P(Bio or Chem) = 0.98 - 0.4

P(Bio or Chem) = 0.58

Therefore, the probability that the student will pass either Biology or Chemistry is 0.58, or 58%.