Kapil and Chirag throw a dice. Find the probability that Kapils throw

Question

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

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 *   = 21/36 = 7/12

stne · Answer

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.

Vivek1707 · Answer

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

shivani1351 · Answer

All possible outcomes of dice = {1,2,3,4,5,6}

We need to find out the probability that Kapil's throw is not greater than, i.e. lesser than or equal to Chirag's throw:

K <= C

1 <= 1 (1 way)
1,2 <= 2 (2 ways)
1,2,3 <= 3 (3 ways)
1,2,3,4 <= 4 (4 ways)
1,2,3,4,5 <= 5 (5 ways)
1,2,3,4,5,6 <= 6 (6 ways)

Hence, the total number of ways in which Kapil’s throw is not greater than Chirag’s throw.= 1+2+3+4+5+6 = (6*7)/2 = 21 and the total number of ways in which both throw the dice = 6*6 = 36

Thus, the overall probability = 21/36 = 7/12 (Answer)

Kapil and Chirag throw a dice. Find the probability that Kapils throw

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Kapil and Chirag throw a dice. Find the probability that Kapils throw

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