Question

Jason and Jenny are out on a lunch date. Jason gives Jenny a bouquet that has 4 roses, 3 carnations, and 4 orchids. They decide that Jenny gets to choose their lunch order if the flower she randomly picks from the bouquet is a carnation or a rose. What is the probability that Jenny gets to choose lunch?

Answer 1

Answer: The probability that Jenny gets to choose lunch =
`(7)/(11)`

Explanation:

Choosing any color flower from the bouquet makes events mutually exclusive.

Now, the total number of flowers in the bouquet= 4+3+4 = 11

Number of rose=4

Probability of choosing a rose P(R)=
`(4)/(11)`

Number of carnations=3

Probability of choosing a carnation P(C)=
`(3)/(11)`

The probability of choosing a rose or a carnation

P(R or C)=P(R)+P(C)=
`(4)/(11)+(3)/(11)=(7)/(11)`

Thus, the probability that Jenny gets to choose lunch =
`(7)/(11)`

Answer 2

7/11

Explanation:

4+3+4=11 and 7 of the flowers are either roses or carnations.