6 people in total so each can be chosen with a probability of 1/6 -> prob of both of them being chosen is 1/6 * 1/5 --> the order in which they can be chosen can vary in two ways hence ans = 1/6 * 1/5 = 1/15
another way of looking at it is that the first person can be chosen in 2/6 ways and then second person can be chosen in 1/5 ways since the first one has already been picked -> again = 1/15
hope that helps.
Joshua and Jose
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life is a test
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heshamelaziry
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life is a test wrote:6 people in total so each can be chosen with a probability of 1/6 -> prob of both of them being chosen is 1/6 * 1/5 --> the order in which they can be chosen can vary in two ways hence ans = 1/6 * 1/5 = 1/15
another way of looking at it is that the first person can be chosen in 2/6 ways and then second person can be chosen in 1/5 ways since the first one has already been picked -> again = 1/15
hope that helps.
Can this be answered in the form of Total desired outcomes/total outcomes ?????????????
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satish.nagdev
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heshamelaziry
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satish.nagdev
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permutations, 6P2 so 6!/(6-2)!
**Edited, to correct mistake its 6P2 not 6P4
**Edited, to correct mistake its 6P2 not 6P4
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heshamelaziry
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heshamelaziry
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Got it. Number of ways of chooisng Jose is 1. Number of ways of choosing Josh is 1. Number of ways choosing any 2 of the 6 = 6! / 2! (6-2)!
Probability = Desired outcomes = 1*1 / 6! / 2! (6-2)! = 1 / 15
Probability = Desired outcomes = 1*1 / 6! / 2! (6-2)! = 1 / 15
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GMATGuruNY
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P(first person selected is Joshua or Jose) = 2/6. (6 total people, among them both Joshua and Jose.)Joshua and Jose work at an auto repair center with 4 other workers. For a survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will both be chosen?
a)1/15
b)1/12
c)1/9
d)1/6
e)1/3
P(next person selected is Joshua or Jose) = 1/5. (5 people left, among them either Joshua or Jose -- but not both -- since one was selected as the first person.)
Since we want these events to happen together, we multiply the fractions:
2/6 * 1/5 = 1/15.
The correct answer is A.
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crisro
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GMATGuruNY wrote,
P(first person selected is Joshua or Jose) = 2/6. (6 total people, among them both Joshua and Jose.)
P(next person selected is Joshua or Jose) = 1/5. (5 people left, among them either Joshua or Jose -- but not both -- since one was selected as the first person.)
Since we want these events to happen together, we multiply the fractions:
2/6 * 1/5 = 1/15.
The correct answer is A.
but,
what if the two persons are selected together and not in two steps?
P(first person selected is Joshua or Jose) = 2/6. (6 total people, among them both Joshua and Jose.)
P(next person selected is Joshua or Jose) = 1/5. (5 people left, among them either Joshua or Jose -- but not both -- since one was selected as the first person.)
Since we want these events to happen together, we multiply the fractions:
2/6 * 1/5 = 1/15.
The correct answer is A.
but,
what if the two persons are selected together and not in two steps?
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GMATGuruNY
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When we're asked to determine the probability that events will happen together, it makes no difference whether the events happen at the same time or one after the other; the probability is the same.crisro wrote:GMATGuruNY wrote,
P(first person selected is Joshua or Jose) = 2/6. (6 total people, among them both Joshua and Jose.)
P(next person selected is Joshua or Jose) = 1/5. (5 people left, among them either Joshua or Jose -- but not both -- since one was selected as the first person.)
Since we want these events to happen together, we multiply the fractions:
2/6 * 1/5 = 1/15.
The correct answer is A.
but,
what if the two persons are selected together and not in two steps?
Another example:
A jar contains 3 red marbles and 2 blue marbles. If two marbles are selected at the same time, what is the probability that both marbles will be red?
To solve, we can determine the probability that the "first" marble is red and that the "second" marble also is red:
P(first marble is red) = 3/5.
P(second marble is red) = 2/4.
Since both events must happen together, we multiply:
3/5 * 2/4 = 3/10.
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@GMATGuruNY,
What if I want to solve the marbles problem with the approach 1-Probability None?
PS: i can't solve the Joshua/Jose problem with this approach because we have more than two choices?
Thank you so much
What if I want to solve the marbles problem with the approach 1-Probability None?
PS: i can't solve the Joshua/Jose problem with this approach because we have more than two choices?
Thank you so much
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GMATGuruNY wrote:P(first person selected is Joshua or Jose) = 2/6. (6 total people, among them both Joshua and Jose.)Joshua and Jose work at an auto repair center with 4 other workers. For a survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will both be chosen?
a)1/15
b)1/12
c)1/9
d)1/6
e)1/3
P(next person selected is Joshua or Jose) = 1/5. (5 people left, among them either Joshua or Jose -- but not both -- since one was selected as the first person.)
Since we want these events to happen together, we multiply the fractions:
2/6 * 1/5 = 1/15.
The correct answer is A.
Hi
can it be considered this way selecting 2 ppl from 6 is 6c2.
probability of selecting only these 2 is 1/6c2
