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29 Jun 2021, 02:45
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
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Question Stats:
66% (02:36) correct
34%
(03:01)
wrong
based on 302
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There are 5 King cards, 6 Queen cards and 7 Jack cards in a box. Rohit draws 3 cards from it, one after another. What is the probability of the second card drawn being a Jack, if the first card drawn is not replaced?
(A) 56/153
(B) 7/18
(C) 60/153
(D) 65/153
(E) 137/306
(A) 56/153
(B) 7/18
(C) 60/153
(D) 65/153
(E) 137/306
sting8
Current Student
Joined: 03 Dec 2020
Last visit: 12 Aug 2023
Posts: 55
Given Kudos: 215
Location: India
Debrief: https://gmatclub.com/forum/740-v40-q50-second-attempt-self-preparation-375379.html
Concentration: Technology, Strategy
Schools: ISB '23 (A) Tepper '24 (WL) Ross '24 (D)
GMAT 1: 720 Q49 V39
GMAT 2: 740 Q50 V40
WE:Engineering (Computer Hardware)
01 Jul 2021, 02:32
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Hi PrajayM
We are picking cards one after other, but once the first card is out, it is not put back into the deck, so this is a case without replacement.
We can pick any card on each of the two picks, only constraint being no replacement. One after other does not mean, after selecting K1,we are bound to pick K2, or only after selecting J3 we can select J4 and so on.
I think this clarifies your doubt.
Since, we have to select the second card as Jack. There are 2 cases :
CASE 1
1st card is not a jack, second card is a jack
P1 = \(\frac{11}{18} * \frac{7}{17}\)
CASE 2
1st card is a jack, second card is also a jack
P2 = \(\frac{7}{18} * \frac{6}{17}\)
Required = P1 + P2 = \(\frac{7}{18}\)
Please consider giving kudos if this helps.
We are picking cards one after other, but once the first card is out, it is not put back into the deck, so this is a case without replacement.
We can pick any card on each of the two picks, only constraint being no replacement. One after other does not mean, after selecting K1,we are bound to pick K2, or only after selecting J3 we can select J4 and so on.
I think this clarifies your doubt.
Since, we have to select the second card as Jack. There are 2 cases :
CASE 1
1st card is not a jack, second card is a jack
P1 = \(\frac{11}{18} * \frac{7}{17}\)
CASE 2
1st card is a jack, second card is also a jack
P2 = \(\frac{7}{18} * \frac{6}{17}\)
Required = P1 + P2 = \(\frac{7}{18}\)
Please consider giving kudos if this helps.
General Discussion
29 Jun 2021, 05:56
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The question is just asking: what is the probability the second card from the top is a Jack? That card is just as likely to be a Jack as any other card in the deck, so the answer is 7/18.
If we had some information about the first card we picked, then that information might influence the probability the next card is a Jack, but we have no information about the first card here, so it's irrelevant that we're even picking it.
If we had some information about the first card we picked, then that information might influence the probability the next card is a Jack, but we have no information about the first card here, so it's irrelevant that we're even picking it.
