Question

A box contains 20 chips. 8 of these chips have a value of $5 and the others have no value. We randomly pick 3 chips from the box and putting them back after each draw.

A = Number of chips of $5 after 2 draws.

What is the distribution of A?

Answer 1

Solution

Back-up Theory

Number of ways of selecting r things out of n things is given by ⁿC_r = (n!)/{(r!)(n - r)!}….....................................…(1)

Values of ⁿC_r can be directly obtained using Excel Function: Math & Trig COMBIN….......................................…. (1a)

Probability of an event E, denoted by P(E) = n/N ……......................................................................…………………(2)

where n = n(E) = Number of outcomes/cases/possibilities favourable to the event E and

N = n(S) = Total number all possible outcomes/cases/possibilities.

Now, to work out the solution,

Vide (1),

Number of ways of selecting 3 chips out of 20 chips = ²⁰C₃ = 1140. [vide (1a)]

So, vide (2), N = 1140 ............................................................................................................................................. (3)

For a single draw of picking 3 chips, let A1 = Number of chips of $5 in the first draw.

A1 = 0 => all 3 chips are $0 chips => Number of ways of selecting 3 chips out of 12 $0 chips = ¹²C₃ = 220. [vide (1a)].

This is n vide (1). Hence , vide (1) and (3), P(A1 = 0) = 220/1140 = 0.1930

A1 = 1 => 1 chips is $5 and 2 are $0 chips => Number of ways of selecting 1 chip out of 8 $5 chips and 2 chips out of 12 $0 chips = (⁸C₁) x (¹²C₂)= 8 x 66. [vide (1a)] = 528. This is n vide (1). Hence , vide (1) and (3), P(A1 = 1) = 528/1140 = 0.4632

A1 = 2 => 2 chips are $5 and 1 is $0 chips => Number of ways of selecting 2 chips out of 8 $5 chips and 1 chip out of 12 $0 chips = (⁸C₂) x (¹²C₁)= 28 x 12. [vide (1a)] = 336. This is n vide (1). Hence , vide (1) and (3), P(A1 = 2) = 336/1140 = 0.2947

A1 = 3 => all 3 chips are $5 chips => Number of ways of selecting 3 chips out of 8 $5 chips = ⁸C₃ = 56. [vide (1a)]

This is n vide (1). Hence , vide (1) and (3), P(A1 = 3) = 56/1140 = 0.0492

If A2 = Number of chips of $5 in the second draw, probability distribution of A2 is identical to that of A1 as given above.

Now, A = A1 + A2 and P(A) = P(A1) x P(A2).

All possible values of A and the corresponding probabilities are given in the following Table.

A1	A2	A	P(A)
0	0	0	0.0372
0	1	1	0.0894
0	2	2	0.0569
0	3	3	0.0095
1	0	1	0.0894
1	1	2	0.2145
1	2	3	0.1365
1	3	4	0.0228
2	0	2	0.0569
2	1	3	0.1365
2	2	4	0.0869
2	3	5	0.0145
3	0	3	0.0095
3	1	4	0.0228
3	2	5	0.0145
3	3	6	0.0024

From the above we derive the probability distribution of A as given below;

A	P(A)
0	0.0372
1	0.1788
2	0.3283
3	0.2920
4	0.1324
5	0.0290
6	0.0024
Total	1.0000

Answer

DONE