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
A) there will be an equal number of males and females at the dinner party?
B) Mandy will be among these invited?
C) there will be only one female guest?
D) they will have only male guests?
E) there will be at least one female guest?
A) To have an equal number of males and females at the dinner party, they need to choose 4 males and 4 females out of their 20 good friends. The total number of ways they can choose 4 males out of 12 is given by the combination formula: C(12, 4) = 12! / (4! * (12-4)!) = 495.
Similarly, the total number of ways they can choose 4 females out of 8 is given by the combination formula: C(8, 4) = 8! / (4! * (8-4)!) = 70.
Therefore, the total number of ways they can have an equal number of males and females is: 495 * 70 = 34,650.
The total number of ways they can invite any 8 friends out of their 20 is given by the combination formula: C(20, 8) = 20! / (8! * (20-8)!) = 125,970.
So, the probability that there will be an equal number of males and females at the dinner party is: 34,650 / 125,970 = 0.2757 (rounded to four decimal places).
B) The probability of Mandy being among the invited guests is 1 out of 20, as there is only one Mandy among their 20 good friends. Therefore, the probability that Mandy will be among these invited is 1 / 20 = 0.05 (rounded to two decimal places).
C) To have only one female guest, they need to choose 7 males and 1 female out of their 20 good friends. The total number of ways they can choose 7 males out of 12 is given by the combination formula: C(12, 7) = 12! / (7! * (12-7)!) = 792.
Similarly, the total number of ways they can choose 1 female out of 8 is given by the combination formula: C(8, 1) = 8! / (1! * (8-1)!) = 8.
Therefore, the total number of ways they can have only one female guest is: 792 * 8 = 6,336.
So, the probability that there will be only one female guest is: 6,336 / 125,970 = 0.0503 (rounded to four decimal places).
D) To have only male guests, they need to choose 8 males out of their 12 male friends. The total number of ways they can choose 8 males out of 12 is given by the combination formula: C(12, 8) = 12! / (8! * (12-8)!) = 495.
Therefore, the total number of ways they can have only male guests is: 495.
So, the probability that they will have only male guests is: 495 / 125,970 = 0.0039 (rounded to four decimal places).
E) The probability of having at least one female guest can be calculated by subtracting the probability of having only male guests from 1. So, the probability of having at least one female guest is: 1 - 0.0039 = 0.9961 (rounded to four decimal places).
This is a test post.
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To solve these probability questions, we need to consider both the total number of possible outcomes and the number of favorable outcomes for each scenario.
Given information:
Total number of friends (N) = 20
Number of male friends (M) = 12
Number of female friends (F) = 8
Number of guests to be invited (G) = 8
A) To have an equal number of males and females at the dinner party, we need to select 4 males and 4 females.
Number of ways to select 4 males out of 12 (M1) = C(12, 4) = 495
Number of ways to select 4 females out of 8 (F1) = C(8, 4) = 70
Total number of favorable outcomes = M1 * F1 = 495 * 70 = 34,650
Total number of possible outcomes = C(N, G) = C(20, 8) = 125,970
Probability of having an equal number of males and females = Total favorable outcomes / Total possible outcomes = 34,650 / 125,970 ≈ 0.275
B) To calculate the probability of Mandy being invited, we need to select 7 guests from the 20 friends, including Mandy.
Number of ways to select Mandy (Mandy) = 1
Number of ways to select 6 other guests (G1) = C(N-1, G-1) = C(19, 7) = 5038
Total number of favorable outcomes = Mandy * G1 = 1 * 5038 = 5038
Probability of Mandy being among the invited guests = Total favorable outcomes / Total possible outcomes = 5038 / 125,970 ≈ 0.04
C) To find the probability of having only one female guest, we need to select 1 female and 7 males.
Number of ways to select 1 female out of 8 (F2) = C(8, 1) = 8
Number of ways to select 7 males out of 12 (M2) = C(12, 7) = 792
Total number of favorable outcomes = F2 * M2 = 8 * 792 = 6336
Probability of having only one female guest = Total favorable outcomes / Total possible outcomes = 6336 / 125,970 ≈ 0.05
D) To calculate the probability of having only male guests, we need to select 8 males out of 12.
Number of ways to select 8 males out of 12 (M3) = C(12, 8) = 495
Total number of favorable outcomes = M3 = 495
Probability of having only male guests = Total favorable outcomes / Total possible outcomes = 495 / 125,970 ≈ 0.004
E) To find the probability of having at least one female guest, we can calculate the complement probability of having only male guests.
Probability of having at least one female guest = 1 - Probability of having only male guests
Probability of having at least one female guest ≈ 1 - 0.004 = 0.996
Therefore,
A) Probability of equal number of males and females ≈ 0.275
B) Probability of Mandy being invited ≈ 0.04
C) Probability of having only one female guest ≈ 0.05
D) Probability of having only male guests ≈ 0.004
E) Probability of having at least one female guest ≈ 0.996