If a baseball player has a batting average of 0.420, what is the probability that the player will get at least 2 hits
in the next four times at bat?
prob(hit) = .42
prob(no hit ) = 1 - .42 = .58
at least 2 hits out of 4 ---> exclude : no hits, 1 hit
prob(no hits)
= C(4,0)(.42)^0 (.58)^4
= .113165
prob(1hit)
= C(4,1)(.42)(.58)^3
= .32779
prob(at least 2 hits)
= 1 - (.113165 + .32779)
= appr .559
or
you could do prob(2hits) + prob(3hits) + prob(4hits)
I bet you will get .559
this is dumb
To find the probability that the baseball player will get at least 2 hits in the next four times at bat, we need to calculate the probability of exactly 2 hits, 3 hits, and 4 hits, and then sum them up.
1. First, we need to find the probability of getting a hit in a single at-bat. Given that the player has a batting average of 0.420, the probability of getting a hit in a single at-bat is 0.420.
2. Next, we need to find the probability of getting exactly 2 hits in four at-bats. We can use the binomial distribution formula: P(X=k) = C(n, k) * p^k * (1-p)^(n-k), where n is the total number of trials, k is the number of successful trials, p is the probability of success in a single trial, and C(n, k) is the number of combinations of n items taken k at a time.
For k=2, n=4 and p=0.420, we have:
P(X=2) = C(4, 2) * (0.420)^2 * (1-0.420)^(4-2)
Using the combination formula C(n, k) = n! / (k! * (n-k)!), we can calculate:
P(X=2) = (4! / (2!(4-2)!)) * (0.420)^2 * (1-0.420)^2
P(X=2) = 6 * (0.1764) * (0.3496)
P(X=2) = 0.3686
3. Next, we need to find the probability of getting exactly 3 hits in four at-bats. Again, using the binomial distribution formula:
For k=3, n=4 and p=0.420, we have:
P(X=3) = C(4, 3) * (0.420)^3 * (1-0.420)^(4-3)
P(X=3) = (4! / (3!(4-3)!)) * (0.420)^3 * (1-0.420)^1
P(X=3) = 4 * (0.07408) * (0.5808)
P(X=3) = 0.1711
4. Finally, we need to find the probability of getting exactly 4 hits in four at-bats:
For k=4, n=4 and p=0.420, we have:
P(X=4) = C(4, 4) * (0.420)^4 * (1-0.420)^(4-4)
P(X=4) = (4! / (4!(4-4)!)) * (0.420)^4 * (1-0.420)^0
P(X=4) = (1) * (0.420^4) * (1)
P(X=4) = 0.07408
5. Finally, we can find the probability of getting at least 2 hits by summing up the probabilities of getting exactly 2, 3, or 4 hits:
P(X >= 2) = P(X=2) + P(X=3) + P(X=4)
P(X >= 2) = 0.3686 + 0.1711 + 0.07408
P(X >= 2) ≈ 0.6138 or 61.38%
Therefore, the probability that the player will get at least 2 hits in the next four times at-bat is approximately 61.38%.
To determine the probability that a baseball player will get at least 2 hits in the next four times at bat, we can use the binomial probability formula. The binomial probability formula is given by:
P(x) = (nCx) * p^x * q^(n-x)
Where:
- P(x) is the probability of getting exactly x hits
- n is the total number of trials or times at bat
- x is the number of hits
- p is the probability of getting a hit in a single trial
- q is the probability of not getting a hit in a single trial, which is equal to 1 - p
In this case, the player has a batting average of 0.420. This means the probability of getting a hit in a single trial (p) is 0.420. The probability of not getting a hit in a single trial (q) is 1 - 0.420 = 0.580.
We want to find the probability of getting at least 2 hits, which means we need to calculate the probability of getting exactly 2, 3, and 4 hits and sum them up.
P(at least 2 hits) = P(2 hits) + P(3 hits) + P(4 hits)
Let's calculate each probability:
P(2 hits) = (4C2) * (0.420)^2 * (0.580)^(4-2)
P(3 hits) = (4C3) * (0.420)^3 * (0.580)^(4-3)
P(4 hits) = (4C4) * (0.420)^4 * (0.580)^(4-4)
We can use the binomial coefficient formula for combinations:
nCx = n! / (x! * (n-x)!)
Plugging in the values and calculating each probability:
P(2 hits) = (4! / (2! * (4-2)!)) * (0.420)^2 * (0.580)^2
P(3 hits) = (4! / (3! * (4-3)!)) * (0.420)^3 * (0.580)^1
P(4 hits) = (4! / (4! * (4-4)!)) * (0.420)^4 * (0.580)^0
After calculating each term, we can sum them up to find the total probability of getting at least 2 hits in the next four times at bat.