M05-31

Question

A flower shop has 2 tulips, 2 roses, 2 daisies, and 2 lilies. If two flowers are chosen at random one after the other without replacement, what is the probability of not selecting exactly two tulips?

A.  \frac{1}{8}  
B.  \frac{1}{7}  
C.  \frac{1}{2}  
D.  \frac{7}{8}  
E.  \frac{27}{28}

Bunuel · Accepted Answer

Official Solution:

A flower shop has 2 tulips, 2 roses, 2 daisies, and 2 lilies. If two flowers are chosen at random one after the other without replacement, what is the probability of not selecting exactly two tulips?

A.  \frac{1}{8}  
B.  \frac{1}{7}  
C.  \frac{1}{2}  
D.  \frac{7}{8}  
E.  \frac{27}{28}

Let's find the probability of the opposite event, which is choosing two tulips, and subtract that from 1. The probability of choosing two tulips is  P(both \ tulips) = \frac{2}{8}*\frac{1}{7} = \frac{1}{28} .
 
Therefore, the probability that not both selected flowers are tulips is  P(not \ both \ are \ tulips) = 1 - P(both \ tulips) = 1 - \frac{1}{28} = \frac{27}{28} .

Answer: E

dmatinho · Answer

Hi Bunel,

I thought : 
Probability of not picking two tulips :
6/8*5/7=5/12

I know I didn't get the correct answer but why do I have to find first the probability of getting both tulips (as you suggest)? 
Could you please explain me ?

Thanks,
Daniela

Bunuel · Answer

Notice that we are allowed to have one tulip, while you are calculating the probability of 0 tulips.

P(not 2 tulips) = P(1 tulip) + P(0 tulips) = 2*2/8*6/7 + 6/8*5/7 = 27/28.

vbbvaibhav · Answer

2 tulips out of 8 flowers can be selected by  1 way .

Total No. of ways 2 flowers can be selected out of 8 flowers is  8c2= 7*8/2 = 28

Probability of selecting 2 tulips out of 8 flowers is  1/28 .

Therefor P(Not Selecting a Tulip)=  1-(1/28) = 28-1/28=27/28

swapnils10 · Answer

Hi Bunnel,

Can we take here for non relevant event (Not selecting the tulip) 6C6

and then find the P = 1- (Non relevant event/relevant event) = 1-(6C6/8C2) = 1-(1/28). Is it correct approach ?

Thanks,

Bunuel · Answer

It should be:

P(no \ two \ tulips) = 1 - P(two \ tulips) = 1 - \frac{the \ number \ of \ ways \ to \ pick \ two \ tulips \ out \ of \ two}{the \ number \ of \ ways \ to \ pick \ two \ flowers \ out \ of \ eight}= 1 - \frac{C^2_2}{C^2_8}=\frac{27}{28} .

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

SudsMeister · Answer

Bunuel ,

I have a doubt w.r.t the total possible outcomes.

While, mathematically it makes sense to take 8C2 = 28 ways as the total, but how would it look like if one were to list down all possible outcomes?

We have 2 Roses, 2 Lilies, 2 Daisies, and 2 Tulips. Total flowers = 2R, 2L, 2D, 2T.

While listing down, I could only list 10 combinations.
RL, RD, RT, RR
LD, LT, LL
DT, DD
TT

Since we are selecting, so order shouldn't matter, and since the 2 Rs / Ls / Ds / Ts are similar to each other, so shouldn't the option R1 L be the same as the option R2 L?

When I counted the 2 similar flowers as separate options, then the outcomes are:

R1 L1  
 R1 D1  
 R1 T1 
 R1 L2 
 R1 D2  
 R1 T2  
 R1 R2 
 
  R2 L1  
 R2 D1  
 R2 T1  
 R2 L2  
 R2 D2 
 R2 T2

L1 T1  
 L1 D1  
 L1 T2  
 L2 D2 
 L1 L2

L2 T1  
 L2 D1  
 L2 T2  
 L2 D2

D1 T1  
 D1 T2  
 D1 D2

D2 T1  
 D2 T2

T1 T2

Here the total comes to 28 and matches the result from the formula. But then the flowers, even though similar, are supposed to be counted separately. Why so?

Bunuel · Answer

Why use 8C2 = 28 as total outcomes?

Because  we are selecting 2 flowers out of 8  distinct  positions .

Even though the flowers  look  alike (e.g., both tulips), they are still are  physically distinct objects .

For example:

Tulip1 ≠ Tulip2, even if both are tulips.

So the set of 8 flowers is: T1, T2, R1, R2, D1, D2, L1, L2, so 8 distinct entities. Choosing any 2 of these gives 8C2 = 28 combinations.

kouassidrissa047 · Answer

I like the solution - it’s helpful.

M05-31

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R1 L1	R1 D1	R1 T1	R1 L2	R1 D2	R1 T2	R1 R2
R2 L1	R2 D1	R2 T1	R2 L2	R2 D2	R2 T2
L1 T1	L1 D1	L1 T2	L2 D2	L1 L2
L2 T1	L2 D1	L2 T2	L2 D2
D1 T1	D1 T2	D1 D2
D2 T1	D2 T2
T1 T2