Find the expectation.

Question

The problem is as follows:

A person draws 3 balls from a bag containing 3 white, 4 red and 5 black balls. He is offered ₹10, ₹5 and ₹2 if he draws 3 balls of the samee colour, 2 balls of the same colour and 1 ball of each colour. Find his expectation.

My approach is:

Total expectation= P(3 balls are of the same colour)*10+ P(2 balls of the same colour)*5+ P(1 ball of each colour)*2.

Now, P(3 balls of same colour)= P(3 White coloured balls) + P(3 Red coloured balls) + P(3 Black coloured balls).

Similarly, calculate P(2 balls of same colour) and P(1 ball of each colour).

Am I right? If possible, please provide a detailed solution. Thanks.

Answer 1

If i correctly understood the situation, then the possible occurrences are respecting one of the three patterns

• XXX, three balls of same color, and the win is 10,
• XXY, possibly in an other order, and the win is 5,
• XYZ, and the win is 2.

(Different letters mark different colors.)

Totally there are 3+4+5 = 12 balls. Let us number them, 1,2,3,...,11,12. 1,2,3 are W(hite), 4,5,6,7 are R(ed), the remaining 8,9,10,11,12 B(lack). There are 12.11.10 possibilities to extract them, getting thus ordered extractions.

• How many times do we get the pattern XXX? There are $$3\cdot 2\cdot 1+4\cdot 3\cdot 2+5\cdot 4\cdot 3=6+24+60=90$$ cases.

• How many times do we get the pattern XXY or XYX or YXX? Three times $XXY$. We split the counting depending on the choice of the two colors $X,Y$. There are $$ \begin{aligned} &3(3\cdot 2\cdot (4+5)+4\cdot 3\cdot (3+5)+5\cdot 4\cdot (3+4)) \\ &\qquad =3(54 + 96 + 140) \\ &\qquad= 3\cdot 290 \\ &\qquad= 870 \end{aligned} $$ cases. (One can get the number of this "rather complicated countinq" after having the two simpler numbers, "excluding" them from the total.)

• Finally, XYZ, we permute to get WRB, there are $3!=6$ permutations, so there are $$3!\cdot 3\cdot 4\cdot 5=6\cdot 60=360$$ cases.

The corresponding probabilities are thus $$ \frac{ 90}{1320}\ ,\qquad \frac{870}{1320}\ ,\qquad \frac{360}{1320}\ . $$ The expected win is $$ 10\cdot \frac{ 90}{1320}+ 5\cdot \frac{870}{1320}+ 2\cdot \frac{360}{1320} = \frac {199}{44} \approx 4.52272727272727\dots\ . $$

Computer check, since it is easy, using sage here:

balls  = [0..11]
colors = 'WWWRRRRBBBBB'
X, XY, XYZ = 0, 0, 0

A = Arrangements( balls, 3 )
for a in A:
    extracted_colors = len( Set( [ colors[a[k]] for k in [0,1,2] ] ) )
    if   extracted_colors == 1:    X   += 1
    elif extracted_colors == 2:    XY  += 1
    elif extracted_colors == 3:    XYZ += 1

print "Probabilities:", X/len(A), XY/len(A), XYZ/len(A)
print "Expected win:", 10*X/len(A) + 5*XY/len(A) + 2* XYZ/len(A)

Results:

Probabilities: 3/44 29/44 3/11
Expected win: 199/44