A bag contains 12 marbles, 7 red and 5 green. The reds are numbered 1-7 and the green are numbered 1-5. If you pick 4 without replacing what is the probability that 2 of the 4 are numbered 5?
2 answers
I think it is 1/11 but don't know how
1/11 is correct.
We can do it in two different ways, easy and difficult.
Easy way:
Pick 2 of the 5-marbles and any of the remainder.
2/12*1/11=2/132=1/66.
The two 5-marbles can be arranged in 4P2 = 4!/(2!2!)= 6 ways, so
probability = 6* (1/66) = 1/11
The difficult way:
We subtract from 1 the cases of picking 1 #5 or none.
P(1-#5)= 4P1 * (1/12*10/11*9/10*8/9)
= 4* (4/33)
= 16/33
P(0-#5)=10/12*9/11*8/10*7/9
= 14/33
P(2-#5) = 1 - (P(1-#5)+P(0-#5)
=1- (30/33)
=3/33
=1/11
Why do we do it the difficult way?
In counting or probability, it is always a good idea to think of different ways of doing the same calculation, to explore different options, and to check our calculations.
We can do it in two different ways, easy and difficult.
Easy way:
Pick 2 of the 5-marbles and any of the remainder.
2/12*1/11=2/132=1/66.
The two 5-marbles can be arranged in 4P2 = 4!/(2!2!)= 6 ways, so
probability = 6* (1/66) = 1/11
The difficult way:
We subtract from 1 the cases of picking 1 #5 or none.
P(1-#5)= 4P1 * (1/12*10/11*9/10*8/9)
= 4* (4/33)
= 16/33
P(0-#5)=10/12*9/11*8/10*7/9
= 14/33
P(2-#5) = 1 - (P(1-#5)+P(0-#5)
=1- (30/33)
=3/33
=1/11
Why do we do it the difficult way?
In counting or probability, it is always a good idea to think of different ways of doing the same calculation, to explore different options, and to check our calculations.
1 answer