A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts two

2 azaleas, 3 buttercups, and 4 petunias for total of 9:
same flower:
2 azaleas- 2/9*1/8 of choosing the same flower.
3 buttercups- 3/9*2/8
4 petunias - 4/9*3/8
2/72+6/72+12/72=20/72 Probability to chhose the same flower.

we want the probability of not choosing so 1-20/72=52/72=26/36=13/18

Let's count the probability of the opposite event and subtract it from 1. Opposite event would be that the florist made a bouquet with two of the same flower: \(\frac{C^2_2+C^2_3+C^2_4}{C^2_{9}}=\frac{10}{36}\) --> \(P=1-\frac{10}{36}=\frac{26}{36}=\frac{13}{18}\)

Answer: B.
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Total flowers : AA, BBB, PPPP

Probability that two same flowers are picked:

AA = 2/9 * 1/8 = 1/36
BB = 3/9 * 2/8 = 1/12
PP = 4/9 * 3/8 = 1/6
--------------------------
1/36 + 1/12 + 1/6 = 5/18

1 - 5/18 = 13/18

Thus, 13/18

Here's my method-

Question asks what is the probability of making bouquet with no two same flowers and we have 2 azaleas, 3 buttercups and 4 petunias
By probability method
1. Probability of Selecting 1 azaleas is 2/9 & Probability of Selecting one of any other flower is 7/8. Therefore 2/9 * 7/8 = 7/36
2. Probability of Selecting 1 buttercup is 3/9 & Probability of Selecting one of any other flower is 6/8. Therefore 3/9 * 6/8 = 1/4
3. Probability of Selecting 1 petunial is 4/9 & Probability of Selecting one of any other flower is 5/8. Therefore 4/9 * 5/8 = 5/18

We can have 1st or 2nd or 3rd combination. Therefore, 7/36 + 1/4 + 5/18 = 13/18

With Combinatric method of probability

total outcomes = 9C2 = 36
selecting either 1 azaleas and 1 buttercup ,or 1 buttercup and 1 petunial, or 1 petunial and 1 azaleas = 2C1*3C1 + 3C1*4C1 + 4C1*2C1 = 26
probability = 26/36 = 13/18

Now reverse combinatric probability

Probability that bouquet is made of 2 same flowers is - either 2 azaleas or 2 buttercup or 2 petunial = 2C2 + 3C2 + 4C2 = 10
total outcomes = 36
Probability that no flowers are same is = 1-10/36 = 26/36 = 13/18


Same answer by all three methods

total possibilities of selecting 2 flowers from 9 = 9*8= 72
position doesn't matter i.e. {AB} & {BA} mean the same. so total cases = \(\frac{72}{2} = 36\)

total cases where both flowers are same.

{AA} = 2*1= 2
{BB} = 3*2= 6
{PP} = 4*3= 12
= 2+6+12= 20

position doesn't matter. so total cases \(= \frac{20}{2} = 10\)

Ans \(= 1 - \frac{10}{36} = \frac{13}{18}\)

required probability = 1- [ probability of both being A'z + probability of both being B'p + probability of both being P'n]

= 1- [ 2/9 * 1/8 + 3/9 * 2/8 + 4/9 * 3/8 ] = 1- 5/18 = 13/18.

Hence B.

Guys, when you use 1-p, why don't you take into account the order and multiply the result, in other words there are 2 azaleas so you should imo

2/9*1/8*2 as you can change order of choosing azalea 1 and 2

I understand that it is wrong from the answer but stlill can't get it

ok.. dint really get your q... can you pls elucidate...

however here is my method... prob(diff flowers) = 1- prob (same flowers)

so prob of having same 2 flowers is . for 2 azaleas can be chosen in 2c2 or 1 way

similarly 2 buttercups can be chosen from 3 in 3c2 or 3 ways

and 2 pertunias from 4 in 4c2 or 6 ways...

therefore total ways is 1+3+6 or 10

total number of possible outcomes is 9c2 or 36

therefore probability of getting the same flowers is 10/36. Probability of not having same flowers is 1-10/36 or 26/36 or 13/18

What we have: 2A, 3B, 4P
Total: 9 flowers

Prob of selecting 2A: (2/9)(1/8)=1/36
Prob of selecting 2B: (3/9)(2/8)=1/12=3/36
Prob of selecting 2P: (4/9)(3/8)=1/6=6/36

1/36+3/36+6/36=10/36=5/18

1-(5/18)=13/18

We can use the following formula:

1 = P(getting two of the same flower) + P(not getting two of the same flower)

Let’s determine the probability of getting two of the same flower.

P(2 azaleas) = 2/9 x 1/8 = 2/72

P(2 buttercups) = 3/9 x 2/8 = 6/72

P(4 petunias) = 4/9 x 3/8 = 12/72

The probability of getting two of the same flower is, therefore, 2/72 + 6/72 + 12/72 = 20/72 = 5/18

Thus, the probability of not getting two of the same flower is 1 - 5/18 = 13/18.

Answer: B
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First, we can rewrite the question as "What is the probability that the two flowers are different colors?"

Well, P(different colors) = 1 - P(same color)

Aside: let A = azalea, let B = buttercup, let P = petunia

P(same color) = P(both A's OR both B's OR both P's)
= P(both A's) + P(both B's) + P(both P's)

Now let's examine each probability:

P(both A's):
We need the 1st flower to be an azalea and the 2nd flower to be an azalea
So, P(both A's) = (2/9)(1/8) = 2/72

P(both B's)
= (3/9)(2/8) = 6/72

P(both P's)
= (4/9)(3/8) = 12/72

So, P(same color) = (2/72) + (6/72) + (12/72) = 20/72 = 5/18

Now back to the beginning:
P(diff colors) = 1 - P(same color)
= 1 - 5/18
= 13/18

Answer: B

Cheers,
Brent

Total ways of selecting 2 flowers out of 9 = 9C2 = 9*8/2! = 36

Favourable = selecting each of different type of flower
= 1 azaleas & 1 buttercup (or) 1 buttercup & 1 petunias (or) 1 azaleas & 1 petunias
= 2C1*3C1 + 3C1*4C1 + 4C1*2C1
= 6 + 12 + 8
= 26

Probability = 26/36 = 13/18

IMO Option B

Pls Hit kudos if you like the solution

Posted from my mobile device

Hi there ,

I solved in this way to find the probability of different two flowers :
(2/9 * 3/8) + (3/9 * 4/8) + (4/9 * 2/8 ) = 13/36

. Plz tell me where I went wrong

Posted from my mobile device

Your solution only accounts for half of the outcomes in which we have two different flowers.
(2/9 * 3/8) represents the probability of selecting an azalea 1st and then a buttercup 2nd, but you don't have the probability of selecting a buttercup 1st and then an azalea 2nd.
Likewise, (3/9 * 4/8) represents the probability of selecting a buttercup 1st and then a petunia 2nd, but you don't have the probability of selecting a petunia 1st and then an buttercup 2nd.
etc.

P(2 Azalea) = 2/9 * 1/8 = 1/36
P(2 Buttercup) = 3/9 * 1/8 = 1/12
P(2 Petunias) = 4/9 * 3/8 = 1/6

1/36 + 1/12 + 1/6 = 10/36

1 - P(two of the same flower) = 1 - 10/36 = 26/36 = 13/18
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P(Not same flower) = {(2c1*3c1)+(2c1*4c1)+(3c1*4c1)}/9c2
= 26/36 = 13/18

Hence, B

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