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

Question

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 a quarter?

(1) The probability the coin will be a nickel or a dime is 1/2.
(2) The probability the coin will be a dime is 1/5.

Archit3110 · Answer

total coins ; n+d+q
find n/n+d+q + q/n+d+q

#1
The probability the coin will be a nickel or a dime is 1/2
n/n+d+q + d /n+d+q = 1/2
not sufficient 
#2
dime ; d/n+d+q= 1/5:0.2
p of nickel or quarter 1-0.2 ; 0.8
IMO B
SUFFICIENT

solikon · Answer

I think that answer should be  B  because if we know the probability of not getting nickel or quarter we got the answer according to formula 1 - result we do not want to get, which is 0.2.

So answer to a question could be find. According to a formula 1 - 0.2 = 0.8.

B

prakashb2497 · Answer

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.

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

chetan2u · Answer

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 ?

prakashb2497 · Answer

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

KarishmaB · Answer

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).

ScottTargetTestPrep · Answer

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).

bumpbot · Answer

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

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