Question

In Braydon’s cafeteria, the meats available are beef and chicken. The fruits available are apple, pear and banana. Braydon is randomly given a lunch with one meat and one fruit. What is the probability that the lunch will include a banana?
A.$\dfrac{1}{3}$
B.$\dfrac{2}{3}$
C.$\dfrac{1}{2}$
D.$\dfrac{1}{5}$
E.$\dfrac{3}{5}$

Answer 1

Hint: We can write all the possible outcomes of giving a lunch with one meat and one fruit using the list of fruits and meets given in the question. Then from that we can find the favourable outcome that banana is included in the lunch. Then we can find the probability by dividing the number of favourable outcomes with the total number of outcomes.

Complete step-by-step answer:
We are given that the meats available are beef and chicken.
The fruits available are apple, pear and banana.
The lunch is given with one meat and one fruit. So, the possible ways of the lunch can be written as,
(beef, apple), (beef, pear) (beef, banana), (chicken, apple), (chicken, pear) and (chicken, banana).
Therefore, there are 6 possible ways of giving lunch with one meat and one fruit.
From the above possibilities, the combination of the lunch with banana are (beef, banana) and (chicken, banana).
Therefore, the number of favourable outcomes is 2.
We know that probability of an event is given by the number of favourable outcomes divided by the total number of possible outcomes.
$P = \dfrac{{no.\,of\,favourable\,outcomes}}{{no.\,of\,possible\,outcomes}}$
On substituting the values, we get,
$P = \dfrac{2}{6}$
On simplification, we get,
$P = \dfrac{1}{3}$
Therefore the required probability is $\dfrac{1}{3}$.
So, the correct answer is option A.

Note: Alternate approach to the problem is given by,
There are 2 meats and 3 fruits available. Therefore, the possible ways of giving lunch with one meat and one fruit is given by,
${}^2{C_1} \times {}^3{C_1} = 2 \times 3 = 6$
As there are only 2 meats available, bananas can be given with any of the 2 meats. So the number of favourable outcomes is 2.
We know that probability of an event is given by the number of favourable outcomes divided by the total number of possible outcomes.
$P = \dfrac{{no.\,of\,favorable\,outcomes}}{{no.\,of\,possible\,outcomes}}$
On substituting the values, we get,
$P = \dfrac{2}{6} = \dfrac{1}{3}$
Therefore the required probability is $\dfrac{1}{3}$.