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Chapter 18: Problem 763

A and B throws a dice. The probability that A wins, if he throws a number higher than \(\mathrm{B}\) is (a) \((1 / 2)\) (b) \((15 / 36)\) (c) \((1 / 36)\) (d) None

Short Answer

Expert verified
The probability that A wins by throwing a higher number than B is \(\frac{5}{12}\), which is not listed in the given options. Therefore, the answer is (d) None.

Step by step solution

01

List all possible outcomes

Since a standard dice has 6 sides, both players A and B have the possibility of rolling numbers 1 through 6 when they throw the dice. Let's represent the outcomes as (A,B) pairs, where A is the number rolled by A and B is the number rolled by B. In this format, there are a total of 6 x 6 = 36 possible pairs of outcomes.
02

Identify winning scenarios for A

To find the probability of A winning the game, we are only interested in outcomes where A rolls a higher number than B. Let's list all these outcomes below: (2,1), (3,1), (3,2), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4), (6,5) In these pairs, we can see that A wins 15 times out of the possible 36 outcomes.
03

Calculate the probability of A winning

We can calculate the probability of A winning by dividing the number of winning outcomes by the total number of outcomes. In this case, the probability would be: Probability = (Number of winning outcomes) / (Total number of outcomes) = 15 / 36
04

Simplify the fraction

Finally, we simplify the fraction to find the final probability: Probability = 15 / 36 = 5 / 12 Since 5 / 12 is not listed in options (a), (b), or (c), the correct answer is: (d) None

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