x, y and z are all unique numbers. If x is chosen randomly

Question

x, y and z are all unique numbers. If x is chosen randomly from the set {7, 8, 9, 10, 11} and y and z are chosen randomly from the set {20, 21, 22, 23}, what is the Probability that x and y are prime and z is not ?

A. 1/5
B. 3/20
C. 13/20
D. 3/10
E. 1/10

A. 1/5
B. 3/20
C. 13/20
D. 3/10
E. 1/10

KarishmaB · Accepted Answer

x from {7, 8, 9, 10, 11} - there are 2 primes here 7 and 11. Probability that x is prime is 2/5.

y and z from {20, 21, 22, 23} - there is only one prime number here 23. Probability that y is prime is 1/4. 
Now you are left with 3 numbers since y and z must be different. You can select z in any way out of those 3 numbers since all are non prime. So probability that z is not prime is 1.

Probability that x and y are prime but z is not = (2/5)*(1/4)*1 = 2/20 = 1/10

Answer (E)

Note that out of y and z, it doesn't matter what you pick first.
If you pick z first, the probability that it is not prime is (3/4). Now you are left with 3 numbers one of which is prime. The probability that y is prime is (1/3). So probability that y is prime and z is not = (3/4)*(1/3) = (1/4) - same as before

nikhilsehgal · Answer

Probability of X to be Prime is: 2/5 as Only two numbers in the list of {7, 8, 9, 10, 11} i.e 7 and 11.

Probability of Y to be Prime is: 1/4 as Only two numbers in the list of {20, 21, 22, 23} i.e 23

Probability of Z not prime is 3/3 because Not prime numbers in the list of {20, 21, 22, 23} are only three.

Hence X*Y*Z = 2/5 *1/4 *3/3 = 1/10

Hope it helps!!

rrsnathan · Answer

Hi, 
Can any one solve this by combination method??

Thanks in Advance,
Rrsnathan

Smallwonder · Answer

Total number of outcomes = (number of ways x can be chosen) * (Number of ways y and z could be chosen)= 5c1 *(4C1*3C1)

Number of ways x is chosen and it is prime = 2C1
Number of ways y is chosen and it is prime = 1C1
Number of ways z is chosen and it is not-prime = 3C1

P = 2C1*1C1*3C1/ 5C1*(4C1*3C1) = 1/10

Hope that helps.

/SW

ankur1901 · Answer

why the probability of Z is 3/3 and not 3/4. I can choose 3 non prime numbers out of the available 4. Should it not be 3/4?

rrsnathan · Answer

Hi,

Y has been picked from the set before Z. So there is only 3 numbers in the list. hence the probablity is Z =3/3

Regards,
Rrsnathan.

Transcendentalist · Answer

The question stem states that x, y and z are unique....

006 · Answer

Just one quick question...how do we know that y is picked up before z...the question says that y & z are picked up randomly...i assumed that they are picked up simultaneously....if we dont know the order, the combination will be (1C1 * 3C1)/4C2 ; Irrespective we get E as the answer (2/5 * 1/4 =1/10 )

jwang27 · Answer

Anyone able to explain this one further?

jwang27 · Answer

Nice! Excellent explanation on picking z vs. y first. Thank you.

Can you please expound on when/whether we have to subtract the probability of the opposite condition? For example, translating the math to a condition: Picking x out of the first set so it is prime, and picking y and z out of the second set such that y is prime, and z is not. Do we not have to subtract the possibility that z is prime, or is that implied given that our condition is X is prime AND y is prime AND z is not prime, and given that z cannot be prime if y is prime? Would we have to subtract the possibility if two prime numbers existed in the second set?

Thanks for the help.

solitaryreaper · Answer

Hi Karishma ,
I have a doubt here.
Even though picking x first or picking y first gives us equal probability i.e 1/4 resulting in probability of 1/10.
 Ideally shouldn't we consider the both cases i.e.picking x first or picking y first while calculating the final probability?
In that case final probability should be 1/10+1/10 = 1/5
Please correct me if I am wrong.

KarishmaB · Answer

It doesn't matter in what way you pick y and z, if the end result is the same, it needs to be considered a single case.

Krystallized · Answer

Hi  VeritasPrepKarishma 
I solved the problem using the following approach:
Total Number of outcomes = 5C1* 4C2  = 30
Number of possible outcomes = 2C1*1*3C1 = 6
Hence the probablity = 6/30 = 1/5.
However, the answer was wrong. I could make out that my mistake was in the highlighted part.

My question is, why can't we choose both y and z together from the set? Why do we need to choose them one-by-one? (i.e. 4C1*3C1)

KarishmaB · Answer

What does 4C2 mean? It means out of 4 elements, 2 are chosen. The 2 are not labelled as first and second. They just form a pair chosen from 4 elements. So they would not be distinct y and z.

But if you want to choose a y value and a z value from 4 numbers, you can choose y in 4 ways and z in 3 ways (or z is 4 ways and y in 3 ways, it doesn't matter).

When you want to just select a pair, you use 4C2. When you want to select two distinct values, you use 4*3.

Hope this is clear.

Krystallized · Answer

Thank you very much  VeritasPrepKarishma  for your explanation. My confusion is clear now..

JeffTargetTestPrep · Answer

Since x is chosen from a set that is different from y and z, we see that the event x is prime is independent from the events y is prime and z is not prime. However, the events y isprime and z is not prime are not independent since they are chosen from the same set. Recall that if two events A and B are independent, then P(A and B) = P(A) x P(B). However, if A and B are not independent, then P(A and B) = P(A) x P(B|A) where P(B|A) means the probability of B given that A has occurred. Thus:

P(x is prime) = 2/5

and

P(y is prime and z is not prime) = P(y is prime) x P(z is not prime, given that y is prime) = 1/4 x 3/3 = 1/4

Finally, we have P(x is prime and (y is prime and z is not prime)) = 2/5 x 1/4 = 2/20 = 1/10.

Answer: E

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x, y and z are all unique numbers. If x is chosen randomly

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