Please awnser thank
1 answer:
You might be interested in
The formula , where r! is 1*2*3*...r
is the formula which gives us the total number of ways of forming groups of r objects, out of n objects.
for example, given 10 objects, there are C(10,6) ways of forming groups of 6, out of the 10 objects.
-----------------------------------------------------------------------------------------------
Selecting 4 people out of 12 can be done in :
many ways.
All the possible groups of 4 people, where the husband and wife are included, can be done in C(10, 2) many ways, since we only calculate the possible choices of 2 out of 10 people, to complete the groups of 4.
Thus, the
probability that both the husband and wife are hired is 45/495=0.09Part 2)
The probability that one is hired and the other is not =
P(husband hired, wife not hired) + P(wife hired, husband not hired)
these 2 are clearly equal, so it is enough to calculate one.
Consider the case : husband hired, wife not hired.
assuming the husband is hired, we have to calculate the possible groups of 3 that can be formed from 11-1 (the wife)=10 people.
this is
thus,
P(husband hired, wife not hired)=120/495=0.24
thus,
The probability that one is hired and the other is not =
P(husband hired, wife not hired) + P(wife hired, husband not hired) =
0.24+0.24=0.48
Answer:
A) 0.09
B) 0.48
Answer:
P(x < 3) = 0.1971
Explanation:
To calculate the probability that x babies weigh more than 20 pounds, we will use the binomial distribution, so it can be calculated as:
So, replacing n = 16 and p = 0.25, we get that the probability is equal to
P(x) = 16Cx (0.25)^x (1 - 0.25)^(16 - x)
Now, the probability that fewer than three weights more than 20 pounds are equal to
Then
P(x < 3) = 0.01 + 0.0535 + 0.1336
P(x < 3) = 0.1971
Therefore, the answer is
P(x < 3) = 0.1971