2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $0.0905153$.

$\endgroup$
1
  • $\begingroup$ Thanks for reply! But I cannot figure out how to get the final answer. Could you please help? Any hint will be appreciated. thanks a lot. $\endgroup$ – hliu Jul 8 '16 at 6:55
-1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.