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You read in a book about bridge that the probability that each of the four players is dealt exactly one ace is about 0.11 This means that (a) in every100 bridge deals, each player has one ace exactly 11 times.

(b) in one million bridge deals, the number of deals on which each player has one ace will be exactly 110,000

(c) in a very large number of bridge deals, the percent of deals on which each player has one ace will be very close to 11%

(d) in a very large number of bridge deals, the average number of aces in a hand will be very close to 0.11

(e) None of these

Short Answer

Expert verified

The part (c) is correct.

Step by step solution

01

Step 1. Given Information     

0.11 is the chance that each of the four players will be dealt precisely one ace. We must find the best solution to the question of what this probability actually signifies.

02

Step 2. Concept Used  

We can't foresee the outcomes of a chance process, yet they have a regular distribution over a large number of repetitions. According to the law of large numbers, the fraction of times a specific event occurs in numerous repetitions approaches a single number. The likelihood of a chance outcome is its long-run relative frequency. A probability is a number between 0(never happens) and 1(happens frequently) (always occurs).

03

Step 3. Explanation    

0It is assumed that each of the four players will be dealt exactly one ace with a chance of 0.11This implies that (c) is correct, because the probability is always close to the percent obtained for very large samples. The (a) section is incorrect because having one ace 10 or 12times is likewise quite possible. The (b)part is incorrect because 109000 deals can be obtained with just one ace. The component (d) is incorrect because the proportion is close to 0 and not the average. And (e) is untrue because (c) is correct.

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Most popular questions from this chapter

Winning at tennis A player serving in tennis has two chances to get a serve into play. If the first serve goes out of bounds, the player serves again. If the second serve is also out, the player loses the point. Here are probabilities based on four years of the Wimbledon Championship:18 P(1st serve in) = 0.59 P(win point | 1st serve in) = 0.73 P(2nd serve in | 1st serve out) = 0.86P(win point | 1st serve out and 2nd serve in) = 0.59

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