Mary hits 60% of her free throw shots in yesterdays game. If Mary takes 7 free throw shots in yesterdays game, what is:
A. The probability she scores 5 of those shots?
B. The probability she scores none of the shots?
C. Her average number of hits, or shots that go on?
D. The standard deviation of marys hits, or shots that go in?
E. The probability she scores at least 6 shots?
To answer these questions, we will use the binomial probability formula and the concept of probability.
The binomial probability formula is given by:
P(x) = C(n,x) * p^x * (1-p)^(n-x)
Where:
- P(x) is the probability of getting exactly x successes
- n is the number of trials or shots taken
- p is the probability of success for each trial or shot
- x is the number of successful shots
A. The probability Mary scores 5 out of 7 shots:
P(5) = C(7,5) * (0.6)^5 * (1-0.6)^(7-5)
= (7! / (5! * (7-5)!)) * (0.6)^5 * (0.4)^2
= 21 * 0.07776 * 0.16
≈ 0.2079
So, the probability that Mary scores 5 out of 7 shots is approximately 0.2079.
B. The probability Mary scores none of the shots:
P(0) = C(7,0) * (0.6)^0 * (1-0.6)^(7-0)
= (7! / (0! * 7!)) * (0.6)^0 * (0.4)^7
= 1 * 1 * 0.0016384
= 0.0016384
So, the probability that Mary scores none of the shots is 0.0016384.
C. Her average number of hits or shots that go in:
The average number of hits can be calculated using the formula:
Mean = n * p
Mean = 7 * 0.6
= 4.2
So, Mary's average number of hits is 4.2.
D. The standard deviation of Mary's hits or shots that go in:
The standard deviation can be calculated using the formula:
Standard Deviation = sqrt(n * p * (1-p))
Standard Deviation = sqrt(7 * 0.6 * (1-0.6))
= sqrt(1.68 * 0.4)
= sqrt(0.672)
≈ 0.8192
So, the standard deviation of Mary's hits is approximately 0.8192.
E. The probability Mary scores at least 6 shots:
To calculate the probability of scoring at least 6 shots, we need to sum up the probabilities of scoring 6, 7 shots:
P(at least 6) = P(6) + P(7)
P(6) = C(7,6) * (0.6)^6 * (1-0.6)^(7-6)
= (7! / (6! * (7-6)!)) * (0.6)^6 * (0.4)^1
= 7 * 0.046656 * 0.4
= 0.13068
P(7) = C(7,7) * (0.6)^7 * (1-0.6)^(7-7)
= (7! / (7! * (7-7)!)) * (0.6)^7 * (0.4)^0
= 1 * 0.0279936 * 1
= 0.0279936
P(at least 6) = 0.13068 + 0.0279936
≈ 0.1587
So, the probability that Mary scores at least 6 shots is approximately 0.1587.