Joe likes to go to an old arcade and play 1010 games of Skeeball every day. On a

Question

Joe likes to go to an old arcade and play

1010

games of Skeeball every day. On average, he wins a prize on

5555%

of the games he plays. Today, however, he won only

33

of the

1010

games he played.

This made Joe think that there must be something wrong with how he is throwing the ball today. However, a friend tells him, "You might be throwing the same way you usually do. People will sometimes have a group of bad games just because of random variation, not because of psychology, or because they are doing anything differently, or any other reason. For someone with your statistics, such random bad days wouldn't even be very rare."

Let's see what Joe's friend means. Suppose that Joe's skill level really hasn't changed, so he still has a

5555%

chance of winning each game he plays, like before. If he plays

1010

games a day, on what percentage of days will he win

33

games or

fewer?

Assume that each game is independent of the others.

Fill in the blank: Joe would win

33

games or fewer on

nothing %

of the days that he plays.

(Round to one decimal place as needed.)

Explanation / Answer

X ~ Binomial (n,p)

Where n = 10, p = 0.55

Binomial probability distribution is

P(X) = nCx px (1-p)n-x

So,

P( X <= 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= 10C0 0.550 0.4510 +10C1 0.551 0.459 +10C2 0.552 0.458 +10C3 0.553 0.457

= 0.102

= 10.2%

Probability that Joe would win 3 games of fewer = 10.2%

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