bburton11 wrote:Joshua and Jose work at an auto repair center with 4 other workers. For a survey on health insurance , 2 of the 6 works will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will be chosen?
1/15
1/12
1/9
1/6
1/3
the answer is 1/15.
Imagine Josh and Jose to be 2 red marbles and the other 4 people to be blue marbles. Place the marbles in a box. You are asked to pick the 2 red marbles in two random draws. For your first draw You have probability of 2/6. For your second you have 1/5. So (1/3 )(1/5)=1/15.
You could also use combination. The combination form would be 2C2/6C2 =1/15 This problem is the similar to someone asking you to pick 2 aces from a deck of cards if there is no replacement. For your first pick , there are only 4 aces and 52 cards so 4/52=1/13. Since the first card was an ace, you only have 3 aces left so your prob for 2nd ace is 3/51, hence 1/13 x 1/17 =1/221=4C2/52C2
The general Formula for picking a given category of things which are independent of things in other categories is as follows:
If you are told to pick two persons Josh and Jose from among 6 persons, you know you can do so in 2 random draws. You have:
P(Jose AND=Intersect Josh)= P(Jose) P(Josh|Jose( this says you know Jose has been taken)) and since they are independent = P(Jose) P(Josh)=2/6 x 1/5
If you were asked to pick four people from among the 6 and you want them to be J P R L you will have:
P(JPRL)=P(J)P(P)P(R)P(L)= 4/6 x 3/5 x 2/4 x 1/3=1/15=4C4/6C4
We got the same result because 6C2=6C4!
Hope this helps.
