Find the indicated probabilities. Then determine if the event is unusual. Explain your reasoning.

The probability of winning a game of rock-paper-scissors is [tex]\frac{1}{3}[/tex]. You play nine games of rock-paper-scissors. Find the probability that the number of games you win is:

(a) Exactly three.
(b) More than seven.
(c) Less than four.

(b) The probability that the number of games you win is more than seven is 0.008. The event is [tex]$\square$[/tex] unusual because it is less than or equal to 0.05.
(Round to three decimal places as needed.)

(c) The probability that the number of games you win is less than four is 0.651. The event is not unusual because it is not less than or equal to 0.05.
(Round to three decimal places as needed.)



Answer :

Sure, let's solve this step-by-step.

We are given the probability of winning a game of rock-paper-scissors as [tex]\( \frac{1}{3} \)[/tex], and we consider a series of 9 games. We will be using the concepts from binomial probability distribution to find the indicated probabilities. The binomial probability formula is:

[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]

Where:
- [tex]\( n \)[/tex] is the number of trials (games), which is 9.
- [tex]\( k \)[/tex] is the number of successful trials (wins).
- [tex]\( p \)[/tex] is the probability of success on an individual trial, which is [tex]\( \frac{1}{3} \)[/tex].
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, also read as "n choose k".

### (a) Probability of winning exactly three games:

To find the probability of winning exactly three games, [tex]\( k = 3 \)[/tex]:

[tex]\[ P(X = 3) = \binom{9}{3} \left(\frac{1}{3}\right)^3 \left(1 - \frac{1}{3}\right)^{9-3} \][/tex]

Using the binomial coefficient and simplifying the expression, we find:

[tex]\[ \boxed{0.273} \][/tex]

This indicates that the probability of winning exactly three games is 0.273.

### (b) Probability of winning more than seven games:

To find the probability of winning more than seven games, we need the sum of the probabilities of winning eight games and nine games:

[tex]\[ P(X > 7) = P(X = 8) + P(X = 9) \][/tex]

Compute each term:

[tex]\[ P(X = 8) = \binom{9}{8} \left(\frac{1}{3}\right)^8 \left(1 - \frac{1}{3}\right)^1 \][/tex]
[tex]\[ P(X = 9) = \binom{9}{9} \left(\frac{1}{3}\right)^9 \left(1 - \frac{1}{3}\right)^0 \][/tex]

Summing these up, we find:

[tex]\[ \boxed{0.001} \][/tex]

Given that this is 0.001, which is less than or equal to 0.05, we can state:

The event is unusual because a probability of 0.001 is less than or equal to 0.05.

### (c) Probability of winning less than four games:

To find the probability of winning less than four games, we sum the probabilities for zero, one, two, and three wins:

[tex]\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \][/tex]

We use the binomial formula for each term and sum them up:

[tex]\[ \boxed{0.650} \][/tex]

Which rounds to:

[tex]\[ \boxed{0.65} \][/tex]

Given that this is 0.65, which is not less than or equal to 0.05, we can state:

The event is not unusual because a probability of 0.65 is not less than or equal to 0.05.

In conclusion, the calculated probabilities are:
- (a) The probability of winning exactly three games is 0.273.
- (b) The probability of winning more than seven games is 0.001, and this event is unusual.
- (c) The probability of winning less than four games is 0.65, and this event is not unusual.