In a hospital 6 deliveries have been occurred. The probability that the baby is female equal 0.48
a. Find the probability distribution of the variable representing the number of female babies.
To find the probability distribution of the variable representing the number of female babies, we need to use the concept of the binomial distribution. The binomial distribution is used to model events that have two possible outcomes (in this case, male or female) and where the probabilities are constant.
The probability distribution of the variable representing the number of female babies can be calculated using the formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
- P(X = k) is the probability of getting exactly k female babies,
- n is the total number of deliveries,
- k is the number of female babies,
- p is the probability of a single delivery resulting in a female baby.
In this case, n = 6 (since there were 6 deliveries) and p = 0.48 (the probability that a baby is female).
To find the probability distribution, we need to calculate P(X = k) for each possible value of k (0, 1, 2, 3, 4, 5, and 6). Let's calculate it for each value:
For k = 0:
P(X = 0) = C(6, 0) * 0.48^0 * (1-0.48)^(6-0) = 1 * 1 * 0.52^6 = 0.107
For k = 1:
P(X = 1) = C(6, 1) * 0.48^1 * (1-0.48)^(6-1) = 6 * 0.48 * 0.52^5 = 0.283
For k = 2:
P(X = 2) = C(6, 2) * 0.48^2 * (1-0.48)^(6-2) = 15 * 0.48^2 * 0.52^4 = 0.319
For k = 3:
P(X = 3) = C(6, 3) * 0.48^3 * (1-0.48)^(6-3) = 20 * 0.48^3 * 0.52^3 = 0.213
For k = 4:
P(X = 4) = C(6, 4) * 0.48^4 * (1-0.48)^(6-4) = 15 * 0.48^4 * 0.52^2 = 0.063
For k = 5:
P(X = 5) = C(6, 5) * 0.48^5 * (1-0.48)^(6-5) = 6 * 0.48^5 * 0.52^1 = 0.011
For k = 6:
P(X = 6) = C(6, 6) * 0.48^6 * (1-0.48)^(6-6) = 1 * 0.48^6 * 0.52^0 = 0.001
So, the probability distribution of the variable representing the number of female babies in this case is:
k | P(X = k)
-------------------
0 | 0.107
1 | 0.283
2 | 0.319
3 | 0.213
4 | 0.063
5 | 0.011
6 | 0.001