In a party,there are 8 men and 12 women A) 10 of them are slected randomly. Find the probability that5men and 5 women are selected (b)All people are randomly aranged in two rows to take a photo. If there are 7 people in the first now, finthe probability that all men or all women are in the first row.

A) To find the probability that 5 men and 5 women are selected, we first need to find the total number of ways to select 10 people from 20 (8 men and 12 women). This can be done using the combination formula C(n, r) = n! / [r!(n-r)!], where n is the total number of items, r is the number of items to choose, and "!" denotes factorial.

The total number of ways to select 10 people from 20 is C(20, 10) = 20! / [10!(20-10)!] = 184,756 ways.

Next, we find the number of ways to select 5 men from 8 and 5 women from 12. This is done by multiplying the number of ways to select 5 men from 8 by the number of ways to select 5 women from 12.

The number of ways to select 5 men from 8 is C(8, 5) = 8! / [5!(8-5)!] = 56 ways. The number of ways to select 5 women from 12 is C(12, 5) = 12! / [5!(12-5)!] = 792 ways.

Multiplying these together gives 56 * 792 = 44,352 ways.

The probability is then the number of desired outcomes (44,352 ways) divided by the total number of outcomes (184,756 ways), which is approximately 0.24.

B) For the second part, we first find the total number of ways to arrange 20 people in two rows with 7 people in the first row. This is done by selecting 7 people to be in the first row, and the rest will automatically be in the second row. The total number of ways is C(20, 7) = 20! / [7!(20-7)!] = 775,20 ways.

Next, we find the number of ways to have all men or all women in the first row. This can happen in two ways: either all 8 men are in the first row, or all 7 of the first 12 women are in the first row.

The number of ways to have all 8 men in the first row is 0, because there are only 7 spots in the first row. The number of ways to have 7 women in the first row is C(12, 7) = 12! / [7!(12-7)!] = 792 ways.

So, the probability is the number of desired outcomes (792 ways) divided by the total number of outcomes (775,20 ways), which is approximately 0.001.

A) To find the probability that 5 men and 5 women are selected, we first need to find the total number of ways to select 10 people out of 20 (8 men and 12 women). This can be done using the combination formula C(n, r) = n! / [r!(n-r)!], where n is the total number of items, r is the number of items to choose, and "!" denotes factorial.

The total number of ways to select 10 people out of 20 is C(20, 10) = 20! / [10!(20-10)!] = 184,756 ways.

Next, we find the number of ways to select 5 men out of 8 and 5 women out of 12. This is done by multiplying the number of ways to select 5 men from 8, C(8, 5) = 8! / [5!(8-5)!] = 56 ways, with the number of ways to select 5 women from 12, C(12, 5) = 12! / [5!(12-5)!] = 792 ways.

So, the number of ways to select 5 men and 5 women is 56 * 792 = 44,352 ways.

The probability is then the number of ways to select 5 men and 5 women divided by the total number of ways to select 10 people, which is 44,352 / 184,756 = 0.24 (rounded to two decimal places).

B) For the second part, we need to find the probability that all men or all women are in the first row.

The total number of ways to arrange 20 people in two rows with 7 in the first row is C(20, 7) = 20! / [7!(20-7)!] = 775,20 ways.

The number of ways to have all men in the first row is C(8, 7) = 8 ways (since there are 8 men and 7 spots in the first row).

The number of ways to have all women in the first row is C(12, 7) = 792 ways (since there are 12 women and 7 spots in the first row).

So, the total number of ways to have all men or all women in the first row is 8 + 792 = 800 ways.

The probability is then 800 / 775,20 = 0.00103 (rounded to five decimal places).

A) To find the probability that 5 men and 5 women are selected, we first need to find the total number of ways to select 10 people out of 20 (8 men and 12 women). This can be done using the combination formula C(n, r) = n! / [r!(n-r)!], where n is the total number of items, r is the number of items to choose, and "!" denotes factorial.

The total number of ways to select 10 people out of 20 is C(20, 10) = 20! / [10!(20-10)!] = 184,756 ways.

Next, we find the number of ways to select 5 men out of 8 and 5 women out of 12. This is done by multiplying the number of ways to select 5 men from 8 (C(8, 5) = 8! / [5!(8-5)!] = 56 ways) by the number of ways to select 5 women from 12 (C(12, 5) = 12! / [5!(12-5)!] = 792 ways).

So, the number of ways to select 5 men and 5 women is 56 * 792 = 44,352 ways.

The probability is then the number of ways to select 5 men and 5 women divided by the total number of ways to select 10 people, which is 44,352 / 184,756 = 0.24 (rounded to two decimal places).

B) For the second part, we need to find the probability that all men or all women are in the first row.

The total number of ways to arrange 20 people in two rows with 7 in the first row is C(20, 7) = 20! / [7!(20-7)!] = 775,20 ways.

The number of ways to have all men in the first row is C(8, 7) = 8 ways (since there are only 8 men).

The number of ways to have all women in the first row is C(12, 7) = 792 ways (since there are 12 women).

So, the total number of ways to have all men or all women in the first row is 8 + 792 = 800 ways.

The probability is then 800 / 775,20 = 0.00103 (rounded to five decimal places).