Is this distribution binomial distribution?

Question

Suppose that three persons (A, B and C) throw at a target. And A throws 10 times with the probability 0.3 to hit the target; and B throws 15 times with the probability 0.2; and C throws 20 times with the probability 0.1. Now Determine the probability that the target will be hit at least 12 times.

My solution is as follows:

For each throw, the probability of hitting the target is

Pr(H=1) = Pr(A)Pr(A Hit) + Pr(B)Pr(B Hit) + Pr(C)Pr(C Hit)

which is

Pr(H=1) = (10/45)*0.3 + (15/45)*0.2 + (20/45)*0.1 = 8/45

So, the throw distribution can be seemed as binomial distribution H ~ Bin(45, 8/45) then can get the answer.

Am I right to consider the target hit variable as a binomial distribution?

---

And please give me a help to get the correct answer, any hints will be appreciated. Thanks.

Answer 1

A random variable $H$ is distributed binomially if it represents the number of successes $H=h$ in $n$ independent and identical trials where $n\ge h$

Hence, $H$ is distributed binomially assuming the independence and identicalness of the throws in which case we have:

$$P(H = h) = \frac{45}{h} p^h (1-p)^{n-h}$$

As for the calculation of $p$, it looks like right but you'll have to clarify what exactly is meant by $P(A)$ and so on. How exactly did you come up with the formula? Did you use law of total probability?

I think

$$p = P(hit) = P(hit | A \ throws)P(A \ throws) + P(hit | B \ throws)P(B \ throws) + P(hit | C \ throws)P(C \ throws)$$

where $P(X \ throws)$ is the proportion of throws of $X$ to the total number of throws assuming independence, identicalness or something

Answer 2

No. To see why, note that the MGF of the sum $W = X_1 + X_2$ of two independent binomial random variables $$X_i \sim \operatorname{Binomial}(n_i, p_i), \quad i = 1, 2,$$ is $$M_W(t) = M_{X_1}(t) M_{X_2}(t) = (1 + (e^t - 1) p_1)^{n_1} (1 + (e^t - 1) p_2)^{n_2}.$$ This is not in general equal to the MGF of a single binomial random variable $Y$ with parameters $n = n_1 + n_2$, $p = (n_1 p_1 + n_2 p_2)/n$, which would be $$M_Y(t) = \left(1 + (e^t - 1)\frac{n_1 p_1 + n_2 p_2}{n_1 + n_2} \right)^{n_1 + n_2},$$ except in the case where $p_1 = p_2$.

---

It is worth noting that the correct exact probability of the event described in the question is $$\frac{10229891531523289867038696518983728647}{119209289550781250000000000000000000000} \approx 0.0858145.$$ However, the probability described by your solution would be around $0.0905153$.