Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Florist 2azaleas, 3 buttercups and 4 petunias [#permalink]
16 Mar 2011, 06:34
This post was BOOKMARKED
80% (04:02) correct
20% (00:54) wrong based on 7 sessions
HideShow timer Statistics
10) A florist has 2 azaleas, 3 buttercups and 4 petunias .She puts two flowers together at random in a bouquet. however, the customer calls and stays that she does not want two of the same flower.What is the probability that the florist does not have to change the bouquet?
Re: Florist 2azaleas, 3 buttercups and 4 petunias [#permalink]
16 Mar 2011, 06:45
You should multiply each probability by 2, since it could be Azalea-Buttercup boquet, or Buttercup-Azalea boquet, so you have corresponding probabilities 1/6, 1/3, 2/9. Actually 2 is a number of possible combinations of 2 objects. So the answer is 1/6+1/3+2/9=(6+12+8)/36=13/18 _________________
If my post is useful for you not be ashamed to KUDO me! Let kudo each other!
A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts two flowers together at random in a bouquet. However, the customer calls and says that she does not want two of the same flower. What is the probability that the florist does not have to change the bouquet?
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
Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...