While playing a certain dice game, Chris wins if the sum of

Probability of winning game = 1- Probability of losing game

Probability of losing game = (Probability of not getting sum 7 in any of the three attempts)

Ways of getting sum 7 = (1,6)(2,5)(3,4)(4,3)(5,2)(6,1)= 6 ways
Total ways of getting outcome on two dice =6*6=36
Probability of getting sum as 7 in any attempt =6/36=1/6
Probability of NOT getting sum as 7 in any attempt = 1-(1/6)= 5/6

Probability of losing game =(5/6)*(5/6)*(5/6)=125/216
I.e. Probability of wining game = 1-(125/216) = 91/216

Answer : Option C

I think question sould be read as follows: Charlie has three rolls with two dices each of them.
Hence every roll minimum score would be 2 (1+1) and maximum 12 (6+6).
Probability of having 7 in the first roll is 1/6 (6 out of 36).
You can follow one of the two ways explained here above. Both works.
Hope it helps.

I am not able to understand above approach. (please elaborate it)

however I have got answer using inclusion exclusion principle.....
Considering 3 rolls of 2 dice
A= event that I get sum equal to 7 in first roll
B= event that I get sum equal to 7 in 2nd roll
C= event that I get sum equal to 7 in 3rd roll

in every roll we have six ways to get sum equal to 7 (i.e. 1,6 or 6,1 and so on)

|A∪B∪C| = |A| + |B| + |C| - |A∩B| - |B∩C| - |A∩C| + |A∩B∩C|
= \(1/6+1/6+1/6-1/6*1/6-1/6*1/6-1/6*1/6+1/6*1/6*1/6\)
= \(91/216\)

IS this way correct?

There are 6 possibilities for the sum to be 7. (1,6) (2,5) (3.4) and the three reverse order pairs.
The probability of a win is 6 over 36 or 1/6 and the probability of a loss is 5/6
Now, there are three possible cases: W or LW or LLW
so, 1/6 + 5/6 * 1/6 + 5/6 * 5/6 * 1/6 = 91/216

The sum of two dice can be 7 in six ways - (1,6) , (2,5), (3,4), (4,3) , (5,2) and (6,1)
Total combinations of the numbers on throwing two dice = 36
Thus probability of getting a sum of 3 = 6/36 = 1/6
Probability of not getting a sum of 7 = 1-1/6 = 5/6

Now Chris has 3 attempts to get the desired sum. If he gets sum 7 in 1st attempt than the result is achieved and the other two attempts are not required -> P1= 1/6

If Chris does not gets sum 7 in 1st attempt, then he needs another attempt and if he gets six then prob -> P2= 5/6*1/6 (P(not getting sum 7 in 1st attempt)* P(get sum 7 in 2nd attempt))

Similarly, a 3rd attempt would be needed if Chris fails to gets sum 7 in previous attempts -> P3 = 5/6*5/6*1/6

Total prob = P1 +P2+ P3 = 1/6 + 5/6*1/6 * 5/6*5/6*1/6 = (36+30+25)*6^3 = 91/126

I understand the solution. But I have a meaning problem:
How many dices could be rolled (max)? Three pairs (6) or three dices (3)

I understand that he rolls the
First dice (from 1 to 6). No thing I can do.
Second dice (1 to 6) = 1/6 of chance of getting a 7.
If the sum is below 7:
Third dice (1 to 6). I dont know how to calculate! =)

My final question is: is this problem ambiguous?

Why dont be consider a 4th option where on none of the 3 rolls he gets a 7? shouldnt we be adding [5][/6]^3 to the 91/216?

oh man..I understood that he has to throw 3 dice..not 3 times....
so i took..
probability that it won't work..
1*5/6 x 5/6 = 25/36
and thus, 11/36 is probability that out of 3 dice, he will get 7..

1 - P(He doesnt win in 3 attempts) = P(at least one = he wins)

1 - (30/36)^3

(30/36)^3 = 5^3*6^3 / 6^3*6^3 = 5^3 / 6^3 = 125/216

1 - 125/216 = 91/216

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