A garden center has seven female and eight male holly plants, but they

Question

A garden center has seven female and eight male holly plants, but they aren’t labeled as such. If a customer randomly purchases three plants, what is the probability of getting both male and female plants?

A. 1/8
B. 1/5
C. 4/5
D. 7/8
E. 8/9

A. 1/8
B. 1/5
C. 4/5
D. 7/8
E. 8/9

MayankSingh · Answer

Ans - C
Both male & female plants = 2M 1F or 1M 2F
Total ways of selecting 3 plants = 15C3=455
Ways of selecting in req combination =7C1*8C2+ 7C2*8C1=364

P(req) = 364/455= 4/5

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Archit3110 · Answer

total P chances 7c2*8c1+8c2*7c1 = 168+196 ; 364
total P for all ; 15c3 ; 455
364/455; 4/5
IMO C

ShukhratJon · Answer

Another way of solving is to find the probability of all three chosen plants to be either male or female. Then subtract it from 1.

Probability (at least one male or female) = 1 - Probability (all male or all female)
 
 P = 1 - \frac{7C3 - 8C3}{15C3}

P = 1 - \frac{1}{5} = \frac{4}{5}

C.

philipssonicare · Answer

It should be 7C3+8C3 in the calculation above, otherwise kudos

JonShukhrat · Answer

Many thanks for pointing out the typo.

ccheryn · Answer

suddenly a strange doubt,

I completely agree upon the above method,

can we solve the same problem by considering as follows,

probablility of one male to be selectedis 1/8, 
prbablity of two female to be seleced 2/7
so probalility is 1/8 * 2/7

or 2 male and 1 female = 2/8 * 1/7

addin both gives 4/8*7 = 1/14. i know somewhere i am wrong...

but i want to arrive at the answer in this approach.

can somebody help?

JonShukhrat · Answer

Hi Cheryn,

Overall we have 15 plants, 8 male and 7 female. So:

1. Probability of choosing 1 female out of 15 is 7/15
2. Probability of choosing 1 male out of 15 is 8/15

As you wrote, we can have two different sets of three plants: either Male+Male+Female or Female+Female+Male

Let’s first choose 2 males and 1 female:

8/15 – first male
7/14 – second male, out of remaining 14 plants
7/13 – first female, out of remaining 13 plants
3 – ways of arranging the set: MMF, MFM, and FMM

So the probability of choosing Male+Male+Female is 8/15*7/14*7/13*3 = 28/65

Now we chose 2 females and 1 male:

7/15 – first female
6/14 – second female, out of remaining 14 plants
8/13 – first male out of remaining 13 plants
3 – ways of arranging the set: FFM, FMF, and MFF

So the probability of choosing Female+Female+Male is 7/15*6/14*8/13*3 = 24/65

Total:  Male+Male+Female + Female+Female+Male = 28/65 + 24/65 = 4/5

Learning this method is helpful to understand the Probability concept in greater detail. However, using combinatorics is much less cumbersome.

ccheryn · Answer

From veritas My tutor , Mr. Dave helped atlast ,

wow i am having helping hands in all directions, as soon i solve that to type here, my friend Jonshukhrat explained the way out.. wow... Man u r .. no words for that , Really thanks jon, u r my gmat angel. 
Thanks u very much for that detail explanation, yes this is what i wanted ... man u r great... 1000 kudos to you if i can. thanks my friend.

my mistake is at confusing the direct probability and counting method

in the former i should consider whole (both male and female together). This is my mistake

but in the counting method which you all people used we can solve individually ( using combination as we are dealing with numerator and denominator separately)

so solution is as follows in the direct probability method

for 1 male and 2 female, there are 3 possible ways we can select them

so it gives 3 * 8/15 * 7/14 * 7/13 ( how we get 3 its by 3!/2! or 3c1 or 3c2 , this is similar to binomial but we are doing without replacement)

similarly for 2 females and 1 male , there are 3 possible ways we can sellect them

so it gives 3 * 7/15 * 6/14 * 8/13

adding both we get

3 * 7 * 8 *( 7 +6) / 15 * 14 * 13
1 * 1 *4 /5 * 1*1 = 4/5

ccheryn · Answer

Hi Jon  JonShukhrat , suddenly i got confused on one more part

1. we are multiplying it by 3 for the possible no of ways to arrange the plants ( mmf, fmm, mfm, ( this means order does matter)

but in the previous method ( 8c2*7c1 + 7c2*8c1/15c3) we are using combination, ( which means order does not matter)yet

though its seems silly , i am yet to feel the nuance i think. missing some logic somewhere.

Could you help me find it out?

2. to get this concept i cooked up a simple problem there are 3 people brad, angelina and penelope

we have to select two people of which one should be male and other should be female whats the probabilitiy ( we can directly conceive it as brad and angelina or brad and penolope so 2/3 ways is the answer)

1st method (1c1 * 2c2 + 2c2 *1c1) / 3c2 = 2/3 ways so 2/3

2nd method

either MF , FM

so MF = 1/3 * 2/2 = 1/3 ( my doubt is what does this 1/3 stands for i got the inner concept 1/3 stands for brad and 2/2 stands for either penelope or angelina)
as second step we multiply by 2 as MF,FM 2 ways... so 1/3 *2 gives 2/3 we got the answer..

but i didnt understand what does MF, FM stand for i mean to conceive 
as in this example first MF stands for brad and either penope or angeline
 second FM stands for either penolope or angelina and brad...

its like doing both ways ( ie we care considering the order)... but order does not matter here... hooo i am confused 
Did u understand my question. where i am going wrong???

i think i have solved this, correct me if i am wrong
the second method is not combination method, 2nd method is the permutation method, so here we are not cancelling out the repetitive probabilities

( which is evident in the denominator, denomiator is 3*2 not 3*2 / 2! so we are consider 6 permutation of taking two people out of 3 people )

so if denominator is of permuatation, we have to keep the numerator in the same order to cancel out the probabilities, so in numerator we are considering the order ie all the probabilities of reverse order also)

hence actually the second method is we are solving the combination from the line of permutation.

AM I RIGHT? did i solve my own question above?

JonShukhrat · Answer

Hi Cheryn, Short answer: the probability of any case is usually depicted through fraction, for example a/b. Numerator a is the number of required cases while denominator b is the number of overall possible cases. One caveat is that both a and b must be calculated as either a permutation or combination, but not as a mix. For example: combination/combination or permutation/permutation is correct. But combination/permutation or permutation/combination is incorrect. In our case, when we multiply individual probabilities 8/15*7/14*7/13*3, the denominator of the resulting fraction would be a permutation x/15*14*13, while the numerator is a combination. That's incorrect. In this case you should either turn the numerator into permutation by multiplying it by 3, or turn the denominator into combination by dividing it by 3. In both cases the result is the same. Long answer: Please take a look at the following link, in which Karishma answers similar question in great detail. Link: https://gmatclub.com/forum/probability- ... l#p1577283

JonShukhrat · Answer

If I got your wording correctly, then the above highlighted part seems to be correct. That's good progress. Bravo Cheryn.

ccheryn · Answer

Yes thanks Jon, if somebody has a friend like u..anybody will progress.

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