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Florist 2azaleas, 3 buttercups and 4 petunias [#permalink]
16 Mar 2011, 05:34
66% (04:37) correct
33% (00:54) wrong based on 3 sessions
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, 05:45
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
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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