Exercise 1
There are two fruit baskets A and B. Basket A has one apple and two oranges. Basket B has one banana and one mango. You randomly pick one fruit from each basket.
(i)
Draw a tree diagram showing all possible pairs of fruits.
(ii)
List the sample space.
(iii)
What is the probability of picking one apple and one banana?
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
A tree diagram keeps the two picks separate. Basket A holds $\displaystyle 3$ fruits (one apple, two oranges) and Basket B holds $\displaystyle 2$ fruits (one banana, one mango). The crucial point is that the two oranges are two different fruits: your hand is equally likely to close on either one. So the tree must give them a branch each - otherwise the apple and "an orange" would look equally likely, and they are not.(i) The tree. Draw the first stage as $\displaystyle 3$ branches for Basket A. From the tip of each of those, draw $\displaystyle 2$ branches for Basket B. That gives \(\displaystyle 3\times 2=6\) endpoints, one for each possible pair. Written out branch by branch:
Each of the $\displaystyle 3$ first-stage branches carries probability \(\displaystyle \tfrac{1}{3}\), and each of the $\displaystyle 2$ second-stage branches carries probability \(\displaystyle \tfrac{1}{2}\).(ii) The sample space. Reading the $\displaystyle 6$ endpoints off the tree:
\[S=\{(A,B),(A,M),(O_1,B),(O_1,M),(O_2,B),(O_2,M)\},\qquad n(S)=6 \]
where \(\displaystyle A\) is the apple, \(\displaystyle O_1,O_2\) the two oranges, \(\displaystyle B\) the banana and \(\displaystyle M\) the mango. These $\displaystyle 6$ outcomes are equally likely, which is what makes them useful.(If you record only the kinds of fruit, the list shrinks to four: \(\displaystyle \{(\text{apple},\text{banana}),(\text{apple},\text{mango}),(\text{orange},\text{banana}),(\text{orange},\text{mango})\}\). But those four are not equally likely - each orange pair covers two branches of the tree and so is twice as likely as the matching apple pair. Never divide by $\displaystyle 4$ here.)(iii) Probability of one apple and one banana. Exactly one of the $\displaystyle 6$ equally likely outcomes is \(\displaystyle (A,B)\):
\[P(\text{apple and banana})=\frac{1}{6} \]Check by multiplying along the branch. The chance of the apple from Basket A is \(\displaystyle \tfrac{1}{3}\) ($\displaystyle 1$ apple out of $\displaystyle 3$ fruits), and the chance of the banana from Basket B is \(\displaystyle \tfrac{1}{2}\) ($\displaystyle 1$ banana out of $\displaystyle 2$). The two picks are from different baskets, so neither affects the other, and the branch probability is their product:
\[\frac{1}{3}\times\frac{1}{2}=\frac{1}{6}\ \checkmark \]
The two methods agree. As a further check, all $\displaystyle 6$ endpoints carry \(\displaystyle \tfrac16\) and \(\displaystyle 6\times\tfrac16=1\).Answers: (i) a tree with $\displaystyle 3$ branches (Apple, Orange $\displaystyle 1$, Orange $\displaystyle 2$) each splitting into $\displaystyle 2$ (Banana, Mango), giving $\displaystyle 6$ endpoints; (ii) \(\displaystyle S=\{(A,B),(A,M),(O_1,B),(O_1,M),(O_2,B),(O_2,M)\}\), \(\displaystyle n(S)=6\); (iii) \(\displaystyle P(\text{apple and banana})=\dfrac{1}{6}\).
| Pick from Basket A | Pick from Basket B | Pair at the end of the branch |
| Apple | Banana | (Apple, Banana) |
| Apple | Mango | (Apple, Mango) |
| Orange $\displaystyle 1$ | Banana | (Orange $\displaystyle 1$, Banana) |
| Orange $\displaystyle 1$ | Mango | (Orange $\displaystyle 1$, Mango) |
| Orange $\displaystyle 2$ | Banana | (Orange $\displaystyle 2$, Banana) |
| Orange $\displaystyle 2$ | Mango | (Orange $\displaystyle 2$, Mango) |