Question

Assume there Is a bag of 30 marbles where 15 are blue, 10 are red, and 5 are black.

A) How many combinations are possible to choose 10 marbles?

B)How many permutations are there to choose 6 marbles where the first marble is blue?

C) What is the probability of getting : blue, red, blue, black?

Answer 1

Solution:

We are given that: a bag contains 30 marbles where 15 are blue, 10 are red, and 5 are black.

Part A) How many combinations are possible to choose 10 marbles?

That is we have to find: how many ways 10 marbles can be chosen from 30 marbles.

That is: ₃₀C₁₀ = ......?

Use following combination formula:

Do all possible cancellations to reduce the calculations.

Here 30 is 3 times 10,

28 is 4 times 7 ,

27 is 3 times 9,

25 is 5 times 5,

24 is 6 times 4,

22 is 2 times 11,

21 is 7 times 3

thus after cancelling these numbers we get:

Part B) How many permutations are there to choose 6 marbles where the first marble is blue?

We are given that: First marble is blue and we have to find number of permutations of 6 marbles.

Thus first marble can be chosen from 15 blue marbles in 15 ways and remaining 5 marbles can be arranged from 29 marbles in ₂₉P₅ ways:

First find ₂₉P₅ by using permutation formula:

Thus total number of permutations in which we can arrange 6 marbles such that first marble is Blue

= 15 X ₂₉P₅

= 15 X 14250600

= 213759000

Part C) What is the probability of getting : blue, red, blue, black?

Thus firs blue marble can be selected from 15 blue marbles in 15C1 ways, red marble can be selected from 10 red marbles in 10C1 ways, Blue marble can be selected from remaining 14 blue marbles in 14C1 ways and black marble can be selected from 5 black marbles in 5C1 ways

and total 4 marbles can be selected from 30C4 ways

Thus we get:

Since nC1 = n

15C1 = 15

10C1 = 10

14C1 = 14

5C1 = 5

and

Thus we get: