Question

Can I get a step by step on this type of problem? this is the only probability problems that I currently struggle with.​

Answer 1

Question 1 : 0.44

Question 2: 0.853

Explanation:

We should use combinatorics to solve this problem

Combinations

The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are not allowed

The formula for finding the number of ways you can pick r items from a set of n items where duplicates are eliminated is given by the combinations formula:

`C(n,r) = (n!)/(( r! (n - r)! ))`

where the ! stands for the factorial of the number

Let's take simple example to illustrate this

Suppose we have 3 books labeled A, B. C and we are asked to randomly pick 2 books in one shot. There are 3 possible picks:

AB, AC, BC

Note that though there are six possible arrangements of 2 items from a set of 3, we have to eliminate duplicates because AB is the same as BA.

In the above case, r = 2, n = 3 so using the formula we get

`C(3,2) = (3!)/(( 2! (3 - 2)! ))\\\\= (3!)/(2! * 1! )\\\\\\= (3 * 2 * 1)/((2 * 1)(1))\\\\\\= 3\\\\`

Now to the actual problem

There are a total of 18 students in the class

Number of ways of picking 3 students is

`C(18,3) = (18!)/(( 3! (18 - 3)! ))\\\\= (18!)/(3! * 15! )\\\\\\=816`

So there are 816 possible ways to pick 3 students from 18 students

There are calculators to perform this gruesome task so I will omit the actual steps and just show the result

Let's consider the first question
Out of the 3 students picked 2 are girls. That means the other one is a boy

Number of ways of picking 2 girls out of 10 is C(10, 2)

C(10, 2) = 45

Then we have to consider the lone boy. Number of ways of picking 1 boy out of 8 is

C(8, 1) = 8

Total number of ways of picking 2 girls and 1 boy

= 45 x 8 = 360

Therefore P(2 girls, 1 boy) :

`(360)/(816) = 0.44`

Second question
To find the number of ways that at least 1 boy can appear in the group is the sum of the number of ways 1, 2 or 3 boys can appear in the group.

Rather than do this, let's approach it in a easier fashion

The probability of at least 1 boy appearing in the group is the complement of no boys on the group, i.e. all volunteers are girls

Number of ways of selecting a volunteer group of all 3 girls:

C(10, 3) = 120

P(no boys in group) = P(all 3 girls in group)

`=(120)/(816) = 0.147`

Therefore P(at least 1 boy) = 1 - P(no boys)

= 1-0.147

= 0.853

Could you please let me know if correct? This is a tough subject area for me too :)

Answer 2

Explanation:

Total students in the class=18

Girls=10

Boys=8

a. Pr(Girl, Girl)= 10/18*9/17=90/306.

b. Pr ( at least one is a boy) = BG* BB* GB*GGB*BBG*BBB*BGG*GBG*GBB*BGB*