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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.
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.
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