There are 15 slate rocks, 20 pumice rocks, and 10 granite...

[I_need_a_700plus]

There are 15 slate rocks, 20 pumice rocks, and 10 granite rocks randomly distributed in a certain field. If 2 rocks are to be chosen at random and without replacement, what is the probability that both rocks will be slate rocks?

a) 1/3
b) 7/22
c) 1/9
d) 7/66
e) 2/45

The correct answer was D - 7/66. Can someone please explain this to me? Thanks!

[kartikeya.payautomata]

in all there are 45 number of rocks including all type of them
15 slate+20 pumice+10 granite=45 rocks

2 rocks can be selected in 45C2 ways=22*45

on other hand to select 2 slate rocks can be selected in 15c2 ways = 7*15

hence probability of slate rock picking=(7*15)/(22*45)=7/66

[agha79]

Formula for probability problems is (Favorable outcomes / Total number of outcomes).

Question says that we will choose 2 rockets at random and we need to know what the probability is of both rocks being "Slate Roks"

In first try we will have (15/45) here (45 = 15 + 20 + 10).

For the second try we will have (14/44) because we have made one selection and we are not replacing the first selection we need to subtract 1 from both numerator and denominator of first try

To get the probability you need to now 15/45 * 14/44 = 7/66. (best way is to simplify the fractions before multiplying 1/3 * 2/22 = 7/66)

I hope this helps

[sharok50]

it's a conditional probability question

P(A &B) = P(A) x P(B/A)

P( 1st one slate and 2nd one slate)= P(1st one Slate) x P(2nd one slate/1st one slate)

= 15/45 x 14/44

=7/66

hope that helps

[geetesht]

it's a conditional probability question

P(A &B) = P(A) x P(B/A)

P( 1st one slate and 2nd one slate)= P(1st one Slate) x P(2nd one slate/1st one slate)

= 15/45 x 14/44

=7/66

hope that helps

Definately the best way to solve such problems is conditional Probability !

Good job , did the same thing got the same answer!

[RonPurewal]

Formula for probability problems is (Favorable outcomes / Total number of outcomes).

Question says that we will choose 2 rockets at random and we need to know what the probability is of both rocks being "Slate Roks"

In first try we will have (15/45) here (45 = 15 + 20 + 10).

For the second try we will have (14/44) because we have made one selection and we are not replacing the first selection we need to subtract 1 from both numerator and denominator of first try

To get the probability you need to now 15/45 * 14/44 = 7/66. (best way is to simplify the fractions before multiplying 1/3 * 2/22 = 7/66)

I hope this helps

this is a good solution.

this is a very standard problem with 2 consecutive probabilities. you should make sure that you can solve it without having to use unnecessary terminology/notation, such as conditional-probability notation.

i.e., you should just be able to think about "probability that the first rock is slate" and "then, probability that the second rock is slate". if you have to give these things formal names such as "p(A)" and "p(B/A)", you are probably spending a lot more time than you should.

for instance, consider the probability that the first four rocks are slate rocks.
this isn't any harder than it is with two rocks; the same pattern is followed. it's (15/45) x (14/44) x (13/43) x (12/42).
...but if you try to write that with this conditional notation, oh dear it's going to be ugly.

[angad.patel]

Can you please explain when one should add the individual probabilities of events versus multiplying them?

i.e. how is one supposed to determine whether -

15/45 + 14/44

vs.

15/45 x 14/44

Aren't only independent events supposed to be multiplied? In this case the second event depends on the first (probability a slate rock will be picked and not replaced, etc.)?

[RonPurewal]

Can you please explain when one should add the individual probabilities of events versus multiplying them?

i.e. how is one supposed to determine whether -

15/45 + 14/44

vs.

15/45 x 14/44

Aren't only independent events supposed to be multiplied? In this case the second event depends on the first (probability a slate rock will be picked and not replaced, etc.)?

yes, but, technically, when you do this sort of "without replacement" thing -- i.e., reducing the value of the denominator according to the items that have already been selected -- you have satisfied the criteria for independence.
in general, however, you can trust that the gmat will not test you on "independent vs. non-independent events", so it won't be productive for you to fuss over that particular definition. instead, just make sure that you know that in cases like this -- in which you have the probabilities of CONSECUTIVE events -- you should just multiply the probabilities.

adding the probabilities is reserved for situations in which the events are mutually exclusive (i.e., there is no overlap between the events), and you wish to find the probability of EITHER of the two situations.
for instance, if the probability of drawing a black sock from a drawer full of socks is 2/5, and the probability of drawing a white sock from that drawer is 1/5, then the probability of drawing a sock that is either white OR black is 3/5.

in this case, it should be clear that you wouldn't want to add the probabilities.
the probability that just the first rock is slate is 15/45, so the probability that BOTH of the first two rocks are slate must be SMALLER than this figure. if you add a probability to this, you'll get a larger probability, which wouldn't make sense.