Probability Problem

[bburton11]

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.

[Stuart@KaplanGMAT]

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

Probability = (# of desired outcomes)/(total # of possibilities)

We're only selecting 2 people, so if we want J&J there's only 1 desired outcome.

For total # of possibilites, we're selecting 2 out of 6, so we use the combinations forumula:

nCk = n!/k!(n-k)!

in which n is the total # of entities and k is the # we're selecting.

So, 6C2 = 6!/2!4! = 6*5*4*3*2/2*4*3*2 = 6*5/2 = 15

Therefore, the final answer is 1/15.

[dtweah]

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.