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
11.
Every handshake takes 2. If no one shakes hands twice with another, there are 10+1 total people shaking hands (11).
Name them 1 thru 11. Make a line and 11 goes down the line and shakes 10 hands and then leaves. 10 then shakes 9 hands and leaves. 8 then shakes 7 hands and leaves. Etc. 2 shakes hands with 1 and leaves. 1 just leaves.
11— 10 hs
10— 9 hs
9— 8 hs
8— 7 hs
7— 6 hs
6— 5 hs
5— 4 hs
4— 3 hs
3— 2 hs
2— 1 hs
1— leaves
Total handshakes 55, total people 11.
The number would be 18.
Half of 18 = 9
Two thirds of 18 = 12
9 + 12 = 21
