Let

I - In-state

O - Out-of-state

Given O/I = 0.25

which means O = 100 and I = 400 (because I + O = 500)

From the given information, we can infer that 25 out-of-state students have not decided major while 75 have decided. But this does not provide any information about the in-state students.

Therefore, D is correct.

In second question, we know that events A and B are independent. This means that the happening or non-happening of A does not affect the probability of B. For getting the answer, you just need to subtract the probability of happening of B from 1.

In the third question, we use the combination formula. Let’s test the first option i.e. 15. If 15 cards were to be used and 2 guesses to be made then the possible number of ways of choosing is 15C2. This equals 15!/ 2! 13! = 105. Now 1/105 corresponds to a probability of less than 1%. Since 15 is the least of all the options, we choose this as the correct answer.

Thanks for the reply but I didnt follow that O = 100 and I = 400 part

I didnt follow the soln

As per the property of independent events

P(A and B)= P(A) x P(B)

so here

P(A and B) = 0.6 x 0.3 = 0.18

As per the question’s requirement

P(A and B) = 0.3 x 0.7 (1-0.3)

thats 0.21

Why did u take 1/105 ? @skygill1

Why is my approach incorrect for this method?

Which?

As per the property of independent events

P(A and B)= P(A) x P(B)

so here

P(A and B) = 0.6 x 0.3 = 0.18

As per the question’s requirement

P(A and B) = 0.3 x 0.7 (1-0.3)

thats 0.21

You have found P(A and B), but that is not what the question is asking. It is asking P(B’|A), that is, the probability that B does not occur *given that* A does occur.

In simple words, p(B occuring) + p(B not occuring)=1. Therefore, to find p(B not occuring)=1-p(B occuring), which will be 1-0.6 = .40