Chapter 14.FOM, Problem 3P

### Algebra and Trigonometry (MindTap ...

4th Edition
James Stewart + 2 others
ISBN: 9781305071742

Chapter
Section

### Algebra and Trigonometry (MindTap ...

4th Edition
James Stewart + 2 others
ISBN: 9781305071742
Textbook Problem

# Dividing a Jackpot A game between two players consists of tossing a coin. Player A gets a point if the coin shows heads, and player B gets a point if it shows tails. The first player to get six points wins an $8 , 000 jackpot. As it happens, the police raid the place when player A has five points and B has three points. After everyone has calmed down, how should the jackpot be divided between the two players? In other words, what is the probability of A winning (and that of B winning) if the game were to continue?The French Mathematician Pascal and Fermat corresponded about this problem, and both came to the same correct calculations (though by very different reasonings). Their friend Roberval disagreed with both of them. He argued that player A has probability 3 4 of winning, because the game can end in the four ways H , T H , T T H , T T T and in three of these, A wins. Roberval’s reasoning was wrong.(a) Continue the game from the point at which it was interrupted, using either a coin or a modeling program. Perform the experiment 80 or more times, and estimate the probability that player A wins.(b) Calculate the probability that player A wins. Compare with your estimate from part (a). To determine (a) To perform: The experiment 80 or more times and estimate the probability that player A wins. Explanation Given: A game between two players consists of tossing a coin. Player A gets a point if the coin shows heads, and player B gets a point if it shows tails. The first player to get six points wins an$8,000 jackpot. As it happens, the police raid the place when player A has five points and B has three points.

The French Mathematician Pascal and Fermat corresponded about this problem, and both came to the same correct calculations (though by very different reasonings). Their friend Roberval disagreed with both of them. He argued that player A has probability 34 of winning, because the game can end in the four ways H, TH, TTH, TTT and in three of these, A wins. Roberval’s reasoning was wrong.

Approach:

The probability of an event E is,

P(E)=n(E)n(S)(1).

Here, n(E) denotes the favorable outcome, n(S) denotes the total outcome

To determine

(b)

To find:

The probability that player A wins is 78

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