Permutation Combination : Rolling Dice

Question 3
A punter offers three choices to you

1. Roll a fair pair of dice, you win if the number if greater than 8
2. Roll a fair pair of dice, you lose if the number is greater than 8. If the number is less than 8, then he tosses a coin, you win if the coin turns "heads"
3. Punter tosses two coins, you win if both return "heads".

Which of these three games would you choose

1. A
2. B
3. C
4. B or C, it does not matter

Correct Answer is B. Choice (B)

Explanatory Answer

One can win game A, if he gets either 9, 10, 11 or 12.

• The number of ways in which one can generate 9 is 4 - {3, 6}, {6, 3}, {4, 5}, {5, 4}
• The number of ways in which one can generate 10 is 3 - {4, 6}, {6, 4}, {5, 5}
• The number of ways in which one can generate 11 is 2 - {6, 5}, {5, 6}
• There is one way by which 12 can be generated {6, 6}
• Total number of combinations possible with a pair of dice is 6*6=36

Therefore, Probability of winning in game A is or .

One can calculate the probability of winning game B as a product of probability of getting less than 8 in a toss and the probability of getting "head" from a coin toss.

Or, the probability of winning game B is or .

Probability of winning in game C is .

Game B is the right answer - choice B.

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AIEEE Math Problems, Practice Questions and Answers : Listed Topicwise

Sets, Relations & Functions		Permutation Combination		Sequences & Series
Limits, Continuity & Differentiability		Statistics & Probability		Trigonometry
Mathematical Reasoning