We are given that a bag contains fruits. The probability of getting a banana and then getting a pear is equivalent of the individual probabilities, that is:
[tex]P(\text{B and P)=P(B)xP(p)}[/tex]
Since he gets the banana first, the probability is:
[tex]P(B)=\frac{1}{3+2+1+4}=\frac{1}{10}[/tex]
Now, the probability of getting a pear is given by the number of pears and the total number of fruits minus the banana, that is:
[tex]P(p)=\frac{4}{3+2+4}=\frac{4}{9}[/tex]
Therefore, the total probability is:
[tex]P(\text{B and P)=}(\frac{1}{10})(\frac{4}{9})=\frac{4}{90}=0.044[/tex]
The probability is 4.4%.
