Probability

[Guest]

A certain junior class class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

(A) 3/40,000
(B) 1/3,600
(C) 9/2,000
(D) 1/60
(E) 1/15

Ans is a ) 3/40000

This has already been answered by MGMAT, but my question is I really dont understand why we do not consider the other scenario. According to me answer should be 3/40000*2, even though that option is not available

I saw a post by Stacey saying that we consider the oher scenario only when it is given that order is important..

http://www.manhattangmat.com/forums/og- ... t1246.html

But I have come across many sums where we consider both scenarios even when nothing is mentioned abt the order. I am not posting the sums coz they are from banned source. So is there any rule abt when to consider both and when not to consider

[Guest]

Below is one example where nothing is mentioned about the order but they have taken both ways.i.e 1/6*1/5*2 =1/15

http://www.manhattangmat.com/forums/pro ... t2561.html

Thanks

[RonPurewal]

This has already been answered by MGMAT, but my question is I really dont understand why we do not consider the other scenario. According to me answer should be 3/40000*2, even though that option is not available

here's an analogy:
there are 26 letters in the alphabet.
this question is sort of like saying: 'if i count all the letters in the alphabet forwards, and then i count them all backwards, i get a total of 52. so why aren't there 52 letters?'
answer:
once you've figured out one legitimate way to count something, you just count it that way. if you count the same things twice in two different ways, then, surprise, you'll calculate twice the number you're supposed to get.

--

in this particular problem, there are two different ways in which you could make the selection:
(1) select a junior, and then select a senior;
(2) select a senior, and then select a junior.
the point is that you only need to figure the number of possibilities in ONE of these ways; they are equivalent. these are just like counting the letters in the alphabet forwards vs. backwards: it doesn't matter which way you do it, as long as you do it.

simpler problem: i have 3 shirts and 4 pairs of pants. how many outfits can i make?
answer is 12, of course. i could calculate this as EITHER 3 x 4 or 4 x 3, but i don't want both.

hth

[Guest]

Ron,

Really sorry to push on this, but I am still confused.. If you you consider the below sum

http://www.manhattangmat.com/forums/pro ... t2561.html

The calculation is done as (1/6*1/5)+(1/6*1/5) = 1/15.. So why are we taking two cases ?

One more example is this where they have cosidered all the cases

http://www.manhattangmat.com/forums/for ... -t716.html


Can you please show me two clear cut examples, one where both cases are considered and the other where only one is considered and why do we do so.... I am usually able to come down to 2 answers for any sum involving probability and then I have to randomly guess between them as I am not sure when to consider different cases and when not to.. Thnx and really value ur help...

[Guest]

Ron,

To make things simple for you I am further clarifying my thought process

Joshua and Jose work at an auto repair center with four other workers. For a survey on a healthcare insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will be both chosen?

A. 1/15
B. 1/12
C. 1/9
D. 1/6
E. 1/3

The way I calculated John and joshua sum:

The probability of selecting john is 1/6 and the probability of selecting joshua is 1/5 so combined probability = 1/6*1/5 = 1/30

But there is a possibilty that joshua is selected first and john second so repeat the above calculation and the answer becomes 1/30+1/30=1/15

The way I calculated Twin brother sum:

Probability of selecting the first brother = 60/1000, probability of selecting his brother = 1/800 so total = 60/1000*1/800= 3/40000

But the twin brother can be selected first and the other brother later so repeat the cycle and you get 3/40000 and the combined probabilty becomes 3/20000


So what is wrong with my thought process? As mentioned above it would close this chapter once and for all if you could please provide 2 examples one considering the order and one without ... Regards






The way I calculated the twin brother sum

1) Probability of selecting one of the brothers

[RonPurewal]

Ron,

To make things simple for you I am further clarifying my thought process

Joshua and Jose work at an auto repair center with four other workers. For a survey on a healthcare insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will be both chosen?

A. 1/15
B. 1/12
C. 1/9
D. 1/6
E. 1/3

The way I calculated John and joshua sum:

The probability of selecting john is 1/6 and the probability of selecting joshua is 1/5 so combined probability = 1/6*1/5 = 1/30

But there is a possibilty that joshua is selected first and john second so repeat the above calculation and the answer becomes 1/30+1/30=1/15

The way I calculated Twin brother sum:

Probability of selecting the first brother = 60/1000, probability of selecting his brother = 1/800 so total = 60/1000*1/800= 3/40000

But the twin brother can be selected first and the other brother later so repeat the cycle and you get 3/40000 and the combined probabilty becomes 3/20000


So what is wrong with my thought process? As mentioned above it would close this chapter once and for all if you could please provide 2 examples one considering the order and one without ... Regards






The way I calculated the twin brother sum

1) Probability of selecting one of the brothers

the problem is that joshua and jose are in the SAME POOL, so that it will be possible to pick joshua 1st and jose 2nd, OR to pick jose 1st and joshua 2nd. this is why you have to double the result in that case.
the brothers are not in the SAME POOL - you're picking a senior first AND THEN a junior (or a junior first and then a senior - pick one of these approaches, but not both).

----

analogy problems:

(1) of 4 shirts, 1 is red, and of 5 pairs of shorts, 1 is red. if a shirt and a pair of shorts are selected at random, what's the probability of picking the red shirt and the red shorts?

(2) of 7 shirts, 2 are red. if 2 shirts are selected at random, what's the probability of picking the 2 red shirts?


solutions:

(1) NOT THE SAME POOL - either you pick a shirt and then a pair of shorts, or vice versa. once you've chosen one of these selection methods, there is only ONE WAY you could get the required combination.
so this is (1/4)(1/5) or (1/5)(1/4) = 1/20.

(2) SAME POOL - you could pick the first red shirt and then the second red shirt, OR vice versa. so you have to double the probability: 2(1/7)(1/6) = 1/21.

hth.

[vikram4689]

(2) SAME POOL - you could pick the first red shirt and then the second red shirt, OR vice versa. so you have to double the probability: 2(1/7)(1/6) = 1/21.

ron,
i agree with all what you mentioned above but i think we should also divide by 2! because both the shirts are indistinguishable. this point can be addressed in 2 ways: i can say we should divide by 2! OR i can say that since shirts are indistinguishable, we should not have multiplied by 2! in the first place. this situation is similar to choosing 2 black cards from a pack of 52 cards, the probability of whose=(26/52)*(25/51)

if there were red and green shirts i.e. distinguishable then we should multiply by 2!.

please let me know your take on this point

[tim]

Ron just pulled the 2 to the front in his calculation. Another way you could do this is to consider the probability of getting a red shirt on the first pick (2/7) and multiply that by the probability of drawing the remaining red shirt from the remaining pool of 6 (1/6)..