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: The probability that Jenny gets to choose lunch =\(\frac{7}{11}\)
Step-by-step 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)=\(\frac{4}{11}\)
Number of carnations=3
Probability of choosing a carnation P(C)=\(\frac{3}{11}\)
The probability of choosing a rose or a carnation
P(R or C)=P(R)+P(C)=\(\frac{4}{11}+\frac{3}{11}=\frac{7}{11}\)
Thus, the probability that Jenny gets to choose lunch =\(\frac{7}{11}\)
Answer:
7/11
Step-by-step explanation:
4+3+4=11 and 7 of the flowers are either roses or carnations.