The probability that event M will not occur is 0.8 and the probability

Can this be solved using a tree?

M does not occur .8 -> R occurs .4 = .4*.8=.32 likelihood
M occurs .2 ->Stop = .20 likelihood
Combine both to get 52/100 = 26/50 = 13/25

Why does this method not work here?

Had similar mistake via doing P(M)*P(not R)+P(not M)*P(R). Thinks it's not right because:

M and R are mutually exclusive events with non-zero probability ==> they are NOT independent ==> P(M) and P(not R) can't be multiplied ==> not right to solve via P(M)*P(not R)+P(not M)*P(R)

To find the probability that either event M or event R will occur, we can use the principle of inclusion-exclusion.

The probability that event M will occur is the complement of the probability that event M will not occur, which is 1 - 0.8 = 0.2.

Similarly, the probability that event R will occur is the complement of the probability that event R will not occur, which is 1 - 0.6 = 0.4.

Since events M and R cannot both occur, the probability that either event M or event R will occur is the sum of their individual probabilities minus the probability that both events occur.

Probability(M or R) = Probability(M) + Probability(R) - Probability(M and R)

Probability(M and R) is 0 because both events cannot occur simultaneously.

Probability(M or R) = Probability(M) + Probability(R) - Probability(M and R) = 0.2 + 0.4 - 0 = 0.6

Hence, the correct answer is (C) 3/5.

Bunuel chetan2u

What would be the answer if we had this instead: "The probability that event M will occur is 0.8, and the probability that event R will occur is 0.6"?

Then, the statement that both m and n cannot occur together will not be true as 0.8+0.6 will become 1.4>1, which is not possible.

So, the answer would depend on P(Both). And the value could be anything from 0.8 to 1.0 depending on P(Both).