The probability of hitting a bird with The probability of hitting a bird with a stone lying on a tree by A is $\frac {1}{4}$ and by B is $\frac {3}{5}$. Draw a probability tree diagram to show the probability of all possible ways of hitting the bird.
I couldn't draw the tree diagram. What is the sample space here..???
 A: I assume $A$ and $B$ are people? Let's go with that.
Imagine $A$ throws a stone; either it hits the bird or not. How do you draw the corresponding tree and branches for these two cases? (Probability tree, that is; not the tree with a bird...)
Next, from the two cases of $A$ (i.e., hits the bird or not) you may now consider the two cases of what happens with $B$ throwing their own stone. Similarly, either the stone thrown by $B$ hits or not.
So your possible outcomes are, I think, neither hits the bird, only $A$ hits the bird, only $B$ hits the bird, or both hit the bird. Although, to be realistic, I would think that if the first thrower hits the bird it would alter the probability that the second does, too. (Maybe their throws coincide?) And to be very realistic, probably one should avoid throwing stones at birds (and glass houses) altogether.
A: P(A) = $\frac{1}{4}$
P(A' = Not hitting) = $\frac{3}{4}$
P(B) = $\frac{3}{5}$
P(B' = Not hitting) = $\frac{2}{5}$
There are total three cases -


*

*None of them hit -


P(A') * P(B') =  $\frac{3}{4} \times \frac{2}{5} =  \frac{6}{20}$


*Either of them hit -


P(A) * P(B') + P(A') * P(B) =  $\frac{1}{4} \times  \frac{2}{5} + \frac{3}{4} \times \frac{3}{5} =  \frac{2}{20} + \frac{9}{20} = \frac{11}{20}$


*Both of them hit -


P(A) * P(B) =  $\frac{1}{4} \times \frac{3}{5} =  \frac{3}{20}$
Sample space = $ \left[ \frac{6}{20}, \frac{11}{20}, \frac{3}{20} \right]$
And you can draw Probability Tree Diagram with the help of this link.
