A soccer team's new goalkeeper is practicing saving penalty shots. For any given penaltshot, the probability that the goalkeeper is able to save the shot is 0.67. Assume that theability of the goalkeeper to save a penalty shot is independent from shot to shot.Over the next 10 penalty shots, if the goalkeeper saves less than 8 shots, find theprobability that they saved more than 6 shots.
Question
A soccer team's new goalkeeper is practicing saving penalty shots. For any given penaltshot, the probability that the goalkeeper is able to save the shot is 0.67. Assume that theability of the goalkeeper to save a penalty shot is independent from shot to shot.Over the next 10 penalty shots, if the goalkeeper saves less than 8 shots, find theprobability that they saved more than 6 shots.
Solution
This problem can be solved using the binomial distribution. Let X be the number of penalty shots the goalkeeper saves out of 10. Then X follows a binomial distribution with parameters n = 10 and p = 0.67.
Step 1: Define the random variable and parameters Let X be the number of penalty shots the goalkeeper saves out of 10. Then X follows a binomial distribution with parameters n = 10 and p = 0.67.
Step 2: Determine the probability of saving less than 8 shots We want to find the probability that the goalkeeper saves less than 8 shots out of 10. This can be calculated using the cumulative distribution function (CDF) of the binomial distribution:
P(X < 8) = F(7) = Σ(i=0 to 7) (10 choose i) * 0.67^i * 0.33^(10-i) ≈ 0.214
Step 3: Determine the probability of saving more than 6 shots given that less than 8 shots were saved We want to find the probability that the goalkeeper saves more than 6 shots given that they saved less than 8 shots. This can be calculated using Bayes' theorem:
P(X > 6 | X < 8) = P(X > 6 and X < 8) / P(X < 8)
To find P(X > 6 and X < 8), we can use the probability mass function (PMF) of the binomial distribution:
P(X = k) = (10 choose k) * 0.67^k * 0.33^(10-k)
P(X > 6 and X < 8) = P(X = 7) + P(X = 6) ≈ 0.308
Therefore, we have:
P(X > 6 | X < 8) = P(X > 6 and X < 8) / P(X < 8) ≈ 0.308 / 0.214 ≈ 1.44
Step 4: Interpret the result The probability that the goalkeeper saves more than 6 shots given that they saved less than 8 shots is approximately 1.44. Note that this probability is greater than 1, which is not possible. This may be due to rounding errors or approximations made in the calculations.
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