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02 Oct 2021, 23:55
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
59% (01:53) correct
41%
(01:50)
wrong
based on 217
sessions
History
Date
Time
Result
Not Attempted Yet
Kapil and Chirag throw a dice. Find the probability that Kapil’s throw is not greater than Chirag’s throw.
A. 7/5
B. 12/7
C. 7/12
D. 7/11
e. None of these
A. 7/5
B. 12/7
C. 7/12
D. 7/11
e. None of these
I am getting the answer as 7/12 :
If Kapil got a 1 in his throw:
the probability of Chirag getting a score greater or equal to 1 is 6/6
If Kapil got a 2 in his throw:
Chirag's probability is 5/6 of beating 2
.
.
.
.
.
If Kapil got a 6 in his throw:
the probability of Chirag getting a score greater or equal to 6 is 1/6
The probability of Kapil getting any of the numbers mentioned above in his throw is 1/6
Therefore, probability that Kapil’s throw is not greater than Chirag’s throw is
1/6 * [1+5/6+4/6+3/6+2/6+1/6] = 21/36 = 7/12
If Kapil got a 1 in his throw:
the probability of Chirag getting a score greater or equal to 1 is 6/6
If Kapil got a 2 in his throw:
Chirag's probability is 5/6 of beating 2
.
.
.
.
.
If Kapil got a 6 in his throw:
the probability of Chirag getting a score greater or equal to 6 is 1/6
The probability of Kapil getting any of the numbers mentioned above in his throw is 1/6
Therefore, probability that Kapil’s throw is not greater than Chirag’s throw is
1/6 * [1+5/6+4/6+3/6+2/6+1/6] = 21/36 = 7/12
11 Oct 2021, 22:07
Kudos
Bookmarks
kapil1995
These are all the cases that favor the above condition
First number Kapil"s outcome , second number Chirag's outcome
\(1 1 , 1 2, 1 3, 1 4 , 1 5, 1 6\)
\(2 2, 2 3, 2 4, 2 5, 2 6\)
\(3 3, 3 4, 3 5, 3 6\)
\(4 4, 4 5, 4 6\)
\(5 5 , 5 6\)
\(6 6\)
Total favorable cases \(= 21\)
Over all total outcomes \(= 36\)
Probability = \(\frac{21}{36} = \frac{7}{12}\)
Ans C
Hope it's clear.
Question requires you to find the prob for the case in which P(Kapil) is < = P(Chirag)
*** Key underlying concept is that probabilities are symmetrical meaning that the prob of Kapil winning = prob of Chirag losing, Hence Total prob - prob of tie = 2* probability of Chirag or Kapil winning
First find the prob of tie
Prob of Kapil getting any value = 6/6
Prob of Chirag getting the same value as Kapil = 1/6 ......a
Hence prob is 1 x 1/6 or 1/6
Total prob - Minus prob of tie = 1 - 1/6 = 5/6
Hence prob of Chirag winning = 5/(6*2) i.e. 5/12 ...... b
Hence total required probability (a+b) = 5/12 + 1/6 = 7/12
*** Key underlying concept is that probabilities are symmetrical meaning that the prob of Kapil winning = prob of Chirag losing, Hence Total prob - prob of tie = 2* probability of Chirag or Kapil winning
First find the prob of tie
Prob of Kapil getting any value = 6/6
Prob of Chirag getting the same value as Kapil = 1/6 ......a
Hence prob is 1 x 1/6 or 1/6
Total prob - Minus prob of tie = 1 - 1/6 = 5/6
Hence prob of Chirag winning = 5/(6*2) i.e. 5/12 ...... b
Hence total required probability (a+b) = 5/12 + 1/6 = 7/12
