Question

Q1) Three shots are fired independently on a target with probability of hitting 0.6 for each.
Find the probability that the target will be hit by:

1. One shot.
2. Two shots
3. Three shots
4. No shot.
5. At least one shot.
6. At least two shots.
7. At most one shot.
8. At most two shots.

Q2) If A and B are two events in a sample space, then show that

P(A|B)P(B)= P(B|A)P(A)

Q3) Three coins are tossed. Find the probability that the outcomes are identical given that there is at least on head.

Answer 1

Q1) Using the binomial distribution we get:
1. \[P(one\ shot)=\left(\begin{matrix}3 \\ 1\end{matrix}\right) \times0.6^{1} \times0.4^{2}\]
2.\[P(two\ shots)=\left(\begin{matrix}3 \\ 2\end{matrix}\right) \times0.6^{2} \times0.4^{1}\]

Answer 2

Q2)

Both sides equal
$$
P(A\cap B)
$$

That completes the proof, but use the definition of conditional probability on each side to fill in the rest:
$$
P(A|B)=\cfrac{P(A\cap B}{P(B)}
$$

Answer 3

If you toss a coin 3 times and they are all identical and you know that one is a head, what can you infer about the other 2?

So now that you know that all three are heads, use the binomial distribution to find the probability of exactly 3 heads in 3 tries as you did in question Q1) part 3 put using p=0.5 instead.

Answer 4

Thanks