The correct answer is choice B. When you are multiplying anything by a value greater than one, the answer you get will always be bigger than what you started with.
The reason lies with in the concept of multiplication.
If you have more than one group of something, you will have more than what you started with.
Situation satisfies the criteria for the use of hypergeometric distribution. Since no replacement is made, binomial distribution is not applicable (probability does not remain constant).
A=number of target wattage bulbs
B=number of non-targeted wattage bulbs
a=number of target wattage bulbs selected
b=number of non-targeted wattage bulbs selected
P(a,b)=C(A,a)*C(B,b)/C(A+B,a+b)
where C(n,x)=combination of x items chosen from n=n!/(x!(n-x)!)
For all following problems,
A+B=4+5+6=15
a+b=3 (selected)
(a) Target wattage = 75W
A=6, B=9, a=2, b=1
P(a,b,A,B)
=P(2,1,6,9)
=C(6,2)*C(9,1)/C(15,3)
=15*9/455
=27/91
(b) target wattage = each of the three
Probability = sum of probabilities of choosing 3 40,60,75-watt bulbs
P(3x40W)+P(3x60W)+P(3x75W)
Case (A,B,a,b)
3x40W (4,11,3,0)
3x60W (5,10,3,0)
3x75W(6,9,3,0)
P(3x40W)+P(3x60W)+P(3x75W)
=C(4,3)*C(11,0)/C(15,3)+C(5,3)*C(10,0)/C(15,3)+C(6,3)*C(9,0)/C(15,3)
=4*1/455+10*1/455+20*1/455
=34/455
Can also be solved by elementary counting, for example, for (a),
P(2x75W)
=C(3,2)*6/15*5/14*9/13
=(3)*6/15*5/14*9/13
=27/91 as before
