Kate and her twin sister

[ffearth]

Here is a problem I found on p.189 of MGMAT Word Translations guide :

"Kate and her twin sister Amy want to be in the same relay-race team. There are 6 girls in the group, and only 4 of them will be placed at random in the team. What is the probability that Kate and Amy will both be on the team?"

The problem is solved using combinatorics in the guide. However, I believe it can be solved using only probability. Here is what I did: Probability of Kate being selected: 4/6, probability of of Amy then being selected = 3/5 . P(K) x P(A) = 4/6 x 3/5 = 2/5, which is the answer provided in the book.

However, I multiplied 2/5 by 2 in order to take into account the case where Amy would be selected before Kate. Thus I got 4/5 as the solution of this problem. Can anyone help me understand what is wrong with my reasoning? Many thanks!

[tim]

your approach explicitly assumes that we answer the question about Kate's selection and then answer the question of Amy's selection. having done this, you cannot then talk about selecting Amy then Kate; this changes everything about the way you initially set up the problem..

[ffearth]

Sorry could you elaborate please?

I thought there were two ways in which the two sisters could be selected: first Kate and then Amy, or first Amy and then Kate.
Because the probability that Amy is selected first and then Kate is selected is also 2/5, I concluded that the solution was 4/5 (2/5 + 2/5). However I don't understand why what I did is wrong.

[tim]

you did not calculate the probability that Kate is selected first. you calculated the probability that Kate is selected. go back and look at what you did.. :)

[ffearth]

What would be the probabilty of Kate being selected first? Isn't it 4/6?

By "first" I mean that she is selected before Amy.

[jnelson0612]

What would be the probabilty of Kate being selected first? Isn't it 4/6?

By "first" I mean that she is selected before Amy.

Envision this . . . six girls are standing in front of you. The selection committee chooses one of them. What is the probability that Kate is the one chosen?

[ffearth]

So did I get lucky by solving it that way? " Probability of Kate being selected: 4/6, probability of of Amy then being selected = 3/5 . P(K) x P(A) = 4/6 x 3/5 = 2/5 "

[tim]

no this wasn't lucky. you did it exactly the way you should have..

[ffearth]

I'm sorry I still don't understand why shouldn't we also take into consideration the case where Amy is selected before Kate.

[tim]

then i'll turn the question back to you: in your method, what do you think you did wrong? why do you think you should double the final answer?

[ffearth]

Because there are two ways the two girls can be selected. Amy before Kate and Kate before Amy..

[tim]

then what you need to know is that based on the way you solved the problem, you do not need to double the final answer. that's the way it is. Jamie and I have repeatedly explained why, and if those explanations don't make sense, the best option for you is to accept the truth at face value and incorporate that into how you do these problems..

[ffearth]

Thank you for your help and patience.

What you both said makes sense actually. I understand that multiplying by 2 is wrong, that the probability of Katy being selected first is 1/6. What I don't understand is what kind of method did I use to get 2/5 and why did it work. You said that I set up the problem in a certain way, but in which way exactly? Is this way of setting up the problem written in the MGMAT guide?

I need to understand in order not to repeat my mistake when I will be in a rush during the test.

Please look at this problem :

"A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?"

Here the answer is from the book "Consider one specific case: blue first, then gren, then red. By the domino-effect rule, the probability of this case is 7 blue/16 total * 5 green/15 total * 4 red/14 total = 7/16 * 5/15 * 4/14 = 1/24.

Now consider another case: green firstm then red, then blue. The probability of this case is 5 green/16 total * 4 red/15 total * 7 blue/14 total = 5/16 * 4/15 * 7/14 = 1/24. Notice that all we have done is swap around the numerators. We get the same final probability! This is no accident; the order in which the balls come out does not matter.

Because the three desired gumballs can come out in any order, there are 3! = 6 different cases. All of these cases must have the same probability. Therefore, the overall probability is 6 * 1/24 = 1/4."

My problem is that I have trouble differentiating between this kind of problems and Amy and Kate's.

[tim]

Okay in your new problem you actually DID find the probability of picking the blue marble first etc. In the first problem, the equivalent step (the probability of picking a Kate FIRST), the probability would be 1/6, but if you were to use this method, you'd also have to account for the probability of Kate being chosen second, third, or fourth, which unnecessarily complicates things. so you used the best method for the first problem. if you want to know where you can find more about these methods, make sure to look them up in the probability chapter rather than the combinatorics chapter..

[ffearth]

I did not find the method I used in the probabilty section of the guide! I still do not understand the method I used to solve the problem ( 4/6 x 3/5 = 2/5).

By the way is the probability of Amy being selected second and Kate being selected third 1/5 x 1/4 = 1/20 ?
I tried to solve the problem by accounting for all the ways that Kate and Amy could be selected (12 ways) but I got a wrong answer :(