Matt and his wife have 20 good friends 12 of them are male and eight of them or female. They decide to have a dinner party but can invite only ate guess they decide to invite their guess by giant in their names from a hat. What is the probability that C) there will be only one female guest? D) they will have only male guests? E) there will be at least one female guest?
Combinations
To solve these problems, we need to use the concept of combinations.
The total number of ways to select 8 guests from a group of 20 is given by the combination formula: C(20, 8) = 20! / (8! * (20-8)!) = 125,970.
To find the probability for each scenario, we need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes.
C) To have only one female guest, we need to select 1 female from the 8 available and 7 males from the remaining 12. The number of favorable outcomes is given by: C(8, 1) * C(12, 7) = 8! / (1! * (8-1)!) * 12! / (7! * (12-7)!) = 8 * 792 = 6,336.
Therefore, the probability of having only one female guest is 6,336 / 125,970 ≈ 0.05 or 5%.
D) To have only male guests, we need to select 8 males from the 12 available. The number of favorable outcomes is given by: C(12, 8) = 12! / (8! * (12-8)!) = 495.
Therefore, the probability of having only male guests is 495 / 125,970 ≈ 0.004 or 0.4%.
E) To have at least one female guest, we use the complement rule. The probability of having no female guest is the same as having only male guests, which is 495 / 125,970 ≈ 0.004 or 0.4%. Therefore, the probability of having at least one female guest is 1 - 0.004 ≈ 0.996 or 99.6%.
To calculate the probabilities, we need to determine the total number of possible combinations for selecting 8 guests out of the 20 friends. We can use the combination formula:
nCr = n! / (r! * (n-r)!)
where n is the total number of friends and r is the number of guests to be selected.
For this scenario: n = 20 and r = 8.
So, the total number of combinations is:
C(20,8) = 20! / (8! * (20-8)!)
= 20! / (8! * 12!)
= 20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 / (8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)
= 125,970
Now we can calculate the desired probabilities:
C) Probability of having only one female guest:
The number of ways to select 1 female guest out of 8 is C(8,1), and the number of ways to select the remaining 7 guests from the 12 males is C(12,7). So the probability is:
P(1 female guest) = (C(8,1) * C(12,7)) / C(20,8)
= (8! / (1! * (8-1)!) * 12! / ((7! * (12-7)!)) / 125,970
= (8 * 792) / 125,970
= 6336 / 125,970
≈ 0.0503
Therefore, the probability of having only one female guest is approximately 0.0503 or 5.03%.
D) Probability of having only male guests:
The number of ways to select 8 males out of the 12 available is C(12,8). So the probability is:
P(only males) = C(12,8) / C(20,8)
= (12! / (8! * (12-8)!)) / 125,970
= 495 / 125,970
≈ 0.0039
Therefore, the probability of having only male guests is approximately 0.0039 or 0.39%.
E) Probability of having at least one female guest:
One way to calculate this probability is by finding the complement, which is the probability of having no female guests, and then subtracting it from 1.
P(no female guests) = C(12,8) / C(20,8) (using the same calculation as in D)
= 495 / 125,970
≈ 0.0039
P(at least one female guest) = 1 - P(no female guests)
= 1 - 0.0039
≈ 0.9961
Therefore, the probability of having at least one female guest is approximately 0.9961 or 99.61%.