x and y are prime numbers and z is not (probability)

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sachin_yadav: Master | Next Rank: 500 Posts; Posts: 212; Joined: Mon Dec 06, 2010 12:52 amFollowed by: 1 member; Location: India; Thanked: 5 times

by sachin_yadav » Tue Dec 14, 2010 10:43 pm

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 numbers and z is not ?

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

Answer E

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Wed Dec 15, 2010 12:47 am

sachin_yadav wrote: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 numbers and z is not ?

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

Answer E

Set of X is {7,8,9,10,11} in total 5 numbers .
Set of y and Z {20,21,22,23} 4 numbers

Total combination are 5 * 4 * 3 = 60

For x we can choose any number from 5 , for y any number from 4 , & for z any number from 3 ( because we have all ready taken a number from a set so only 3 numbers are left to choose from )

2 prime numbers are in x , 1 for y & any of the 3 for z

2 * 1 * 3 = 6

6/60 = 1/10

Saurabh Goyal
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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Wed Dec 15, 2010 12:56 am

sachin_yadav wrote: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 numbers and z is not ?

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

Answer E

What Goyalsau has stated is absolutely correct. We can do it in Probability way also.

P(Prime number from set X) = 2/5 As 7 or 11 can be chosen as Prime from {7,8,9,10,11}

P(Prime number for Y) = 1/4 As y can be only 23 from {20,21,22,23}

P(Non Prime for Z) = 3/3 (after selecting for y we have 3 numbers remaining in the set and all are not Primes)

Total Probability = 2/5 * 1/4 * 3/3 = 1/10

If the problem is Easy Respect it, if the problem is tough Attack it

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sachin_yadav: Master | Next Rank: 500 Posts; Posts: 212; Joined: Mon Dec 06, 2010 12:52 amFollowed by: 1 member; Location: India; Thanked: 5 times

by sachin_yadav » Wed Dec 15, 2010 3:20 am

Thank you so much shovan and Goyalsau for clearing my doubt. Both the methods are very good and easy.

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ikokurin: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Fri Apr 06, 2012 6:09 am

by ikokurin » Sat Jun 23, 2012 7:52 pm

Please explain why not 1/5. Here is how I get there:

P(x prime): 2/5
P (y prime/z not prime): 1/4*3/3+3/4*1/3 = 1/2 (i.e. we are satisfied with both situations when: y is prime and z not prime as well as z not prime and y is prime) - I do not understand why you guys are not counting the second scenario.
So the answer is: 2/5*1/2 = 1/5

Hope to hear your feedback.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8137; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Wed Jan 03, 2018 4:15 pm

sachin_yadav wrote: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 numbers and z is not ?

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

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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