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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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.
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Many thanks for pointing out the typo.
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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.
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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?
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If I got your wording correctly, then the above highlighted part seems to be correct. That's good progress. Bravo Cheryn.
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