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?
Combinations
A) To calculate the probability of having an equal number of males and females at the dinner party, we need to consider the possible combinations.
There are 12 male friends and 8 female friends, so the total number of possible combinations is C(12, 4) * C(8, 4) = 495 * 70 = 34,650.
To have an equal number of males and females, they need to invite 4 males and 4 females. The number of combinations to choose 4 males out of 12 is C(12, 4) = 495, and the number of combinations to choose 4 females out of 8 is C(8, 4) = 70.
Therefore, the probability of having an equal number of males and females is 495 * 70 / 34,650 ≈ 0.104 or 10.4%.
B) To calculate the probability of Mandy being among the invited guests, we need to consider the total number of possible combinations.
Since they can invite any 8 guests out of the 20 friends, the total number of possible combinations is C(20, 8) = 125,970.
To calculate the number of combinations where Mandy is invited, we need to consider the number of combinations without her (C(19, 8)) and subtract it from the total number of combinations.
So, the number of combinations with Mandy is C(20, 8) - C(19, 8) = 125,970 - 19,110 = 106,860.
Therefore, the probability of Mandy being among the invited guests is 106,860 / 125,970 ≈ 0.848 or 84.8%.
C) To calculate the probability of having only one female guest, we need to consider the possible combinations.
They can choose 8 guests out of the 20 friends, and they need only one of them to be female. The number of combinations to choose 1 female out of 8 is C(8, 1) = 8.
To choose the remaining 7 guests out of the 19 remaining friends, the number of combinations is C(19, 7) = 26,334.
Therefore, the probability of having only one female guest is 8 * 26,334 / C(20, 8) = 210,672 / 125,970 ≈ 1.672 or 167.2%.
D) To calculate the probability of having only male guests, we need to consider the possible combinations.
They can choose 8 guests out of the 20 friends, and they need all of them to be male. The number of combinations to choose 8 males out of 12 is C(12, 8) = 495.
Therefore, the probability of having only male guests is 495 / C(20, 8) = 495 / 125,970 ≈ 0.004 or 0.4%.
E) To calculate the probability of having at least one female guest, we need to consider the possible combinations.
They can choose 8 guests out of the 20 friends, and we need to subtract the combinations with only male guests.
The number of combinations with only male guests is 495 (as calculated in part D).
Therefore, the probability of having at least one female guest is (C(20, 8) - 495) / C(20, 8) = (125,970 - 495) / 125,970 ≈ 0.996 or 99.6%.
To find the probabilities for each scenario, we first need to calculate the total number of possible outcomes.
Given that Matt and his wife have 20 friends (12 males and 8 females):
A) To have an equal number of males and females at the dinner party, we need to select 4 males and 4 females. The number of ways to do this is given by the combination formula:
C(12, 4) * C(8, 4) = 495 * 70 = 34,650
B) To find the probability that Mandy will be among the invited guests, we need to select Mandy and any 7 other friends. The number of ways to do this is given by the combination formula:
C(1, 1) * C(19, 7) = 1 * 48,620 = 48,620
C) To have only one female guest, we need to select 7 males and 1 female. The number of ways to do this is given by the combination formula:
C(12, 7) * C(8, 1) = 792 * 8 = 6,336
D) To have only male guests, we need to select 8 males. The number of ways to do this is given by the combination formula:
C(12, 8) = 495
E) To have at least one female guest, we need to consider all scenarios except having only male guests (as calculated in question D). So, the total possible outcomes for having at least one female guest is:
Total outcomes - Outcomes with only male guests = C(20, 8) - C(12, 8) = 125,970 - 495 = 125,475
Now, we can calculate the probabilities for each scenario by dividing the number of desired outcomes by the total number of possible outcomes:
A) Probability of having an equal number of males and females = 34,650 / 125,970 ≈ 0.275
B) Probability of Mandy being invited = 48,620 / 125,970 ≈ 0.386
C) Probability of having only one female guest = 6,336 / 125,970 ≈ 0.05
D) Probability of having only male guests = 495 / 125,970 ≈ 0.004
E) Probability of having at least one female guest = 125,475 / 125,970 ≈ 0.995
Note: The probabilities are approximations rounded to three decimal places.