A jar contains only nickels, dimes, and quarters. If a coin is drawn

Would the answer still be B if Question Stem read 'and' instead of 'or'? A jar contains only nickels, dimes, and quarters. If a coin is drawn from the jar at random, what is the probability that it will be either a nickel or and a quarter?

Could we still subtract it from 1 in S2 to get the same answer?

I'm a little confused here, any links to this would be of help.

chetan2u Bunuel KarishmaB ScottTargetTestPrep

Hey
You may be wanting to ask something else.
You can’t change OR to AND. It will not make any sense.
How can one coin be both a nickel and a quarter ?

Oh. Fair enough, when you look at it logically, my question doesn't make sense. Thanks for pointing it out.

If there were P of two events occuring together, P (A & B), would subtracting 1-P(A & B) yield P of (C)?

Evidently, from the above question, 1-P(A or B)= P (C), so I know the same won't be for P(A & B) but I don't fully understand the logic and there's some gap.

Could you help me understand this or point to relevant resources to bridge the gap?

Thanks in advance chetan2u

I am not sure what your question is but this is the relation between or and and:

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

We subtract P(A and B) because it is counted twice - once in P(A) and then in P(B).

In this question, the events A (nickels), B (dimes), and C (quarters) are mutually exclusive and collectively exhaustive. Mutually exclusive means those events cannot happen at the same time, which implies P(A or B or C) = P(A) + P(B) + P(C). For events which are not mutually exclusive, the formula for P(A or B or C) is much more complicated. Collectively exhaustive means it is not possible to observe any outcome other than these three events.

The reason we get P(C) when we subtract 1 - P(A or B) has to with these properties. Since the three events are collectively exhaustive, P(A or B or C) = 1. Since they are mutually exclusive, P(A or B or C) = P(A) + P(B) + P(C), which in turn means P(A) + P(B) + P(C) = 1. From this equation, you can see that 1 - P(A) - P(B) = 1 - P(A or B) = P(C). It is also true that 1 - P(C) = P(A) + P(B) = P(A or B).

This would no longer be true if you had another coin, say a penny, in the jar. If D is the even that the randomly drawn coin is a penny, then 1 - P(A or B) is no longer equal to P(C), but it is equal to P(C or D). Also, 1 - P(C) is no longer equal to P(A or B), but it is equal to P(A or B or D).