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28 Aug 2020, 02:41
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
84% (01:23) correct
16%
(02:06)
wrong
based on 74
sessions
History
Date
Time
Result
Not Attempted Yet
A bag contains 185 cards: 160 “Try Again” cards and 25 “Winner” cards. How many “Try Again” cards must be removed for the probability of selecting a “Winner” card among 25 such cards to be 1/6?
(A) 35
(B) 40
(C) 42
(D) 48
(E) 64
(A) 35
(B) 40
(C) 42
(D) 48
(E) 64
yashikaaggarwal
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 29 Mar 2026
Posts: 3,089
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Location: India
Schools: Cranfield (A) Warwick MiM "23 (M)
GPA: 4
WE:Analyst (Internet and New Media)
28 Aug 2020, 02:56
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The no. of "Try Again" cards to be removed such that \(\frac{Winner}{Winner + Try Again}\) cards become\(\frac{1}{6}\)
\(\frac{Winner}{Winner + Try Again}\)= \(\frac{1}{6}\)
\(\frac{25}{25 + Try Again cards}\) = \(\frac{1}{6}\)
150 = 25 + Try Again cards
125 = Try Again cards
Existing Try Again Cards = 160
Needed Try Again Cards = 125
Try Again Cards to be removed = 160 - 125 = 35
Answer is A
\(\frac{Winner}{Winner + Try Again}\)= \(\frac{1}{6}\)
\(\frac{25}{25 + Try Again cards}\) = \(\frac{1}{6}\)
150 = 25 + Try Again cards
125 = Try Again cards
Existing Try Again Cards = 160
Needed Try Again Cards = 125
Try Again Cards to be removed = 160 - 125 = 35
Answer is A
28 Aug 2020, 03:07
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Quote:
For the probability to be 1/6th, 25 must be 1/6th of the total number of cards, i.e., total number of cards including the winner cards has to be 6*25 = 150.
Given the total number of cards - 185
Required Total number of cards - 150
Cards of Try Again to be removed - 35. A is the answer
