Question

A bag contains 4 Blue, 5 red and 3 green balls. Debolina is asked to pick up three balls at random, one by one without replacement from that bag. If the balls picked up are of same colour then Debolina will get Rs. 7000, if the balls picked up are of different colour, then she will get Rs. 9000, but otherwise she has to pay Rs. 4000.

a) Find the expected gain of Debolina if she plays the game.

b) What is the risk on the expected gain of her?

c) Based on the risk and gain, should she play the game?

Answer 1

Total Number of balls = 4 +5 +3 = 12

Since three balls are drawn without replacement.

Without restriction the total number of ways to draw 3 balls from 12 balls without replacement is

=¹²C₃ = 220

The number of ways to draw to three balls of same colours

= ( Draw three balls of blue colours) or ( Draw three balls of red colours) or ( Draw three balls of green colours)

= ⁴C₃ + ⁵C₃ + ³C₃ = 4 + 10 + 1 =15

The number of ways to draw to three balls of different colours

= Draw 1 blue ball, draw 1 red ball and draw 1 green ball.

= ⁴C₁^{* 5}C₁* ³C₁ = 60

And the remaining ways are = 220 - 15 -60 = 145

If Debolina draw three balls of same colour then she will get 7000 and draw three balls of diffrent colours she will get 9000 otherwise she has to pay Rs . 4000.

Let random variable X denotes amount of gain of Debolina if she has plays the game.

Hence the X takes values 7000,9000,-4000.

The probability distribution of X is

x	-4000	7000	9000	Total
P(X=x)	145/220	15/220	60/220	1
x*p(x)	-580000/220	105000/220	540000/220	295.4545

a) Expected gain of Debolina if she plays the game = E(X)

= 295.4545

b) Risk of expected gain of her = 15/220 +60 /220 = 0.3409

c) Since risk of gain is 0.3409 and expected gain is 295.4545 (Positive) .

Hence Debolina should play the game.