The bombing planes are intersecting two lines of anti-aircraft defense. Each plane, regardless of the others, has probability $\theta$, true resemblance, and can be knocked down by the first line of defense and second line of defense. Probability $\theta$ is not known. Of the n = 100 aircraft, $K_1$ = 40 were knocked down and $K_2$ = 20 were knocked down.
- Compute likelihood for $K_1$ and $K_2$, $P_{\theta}(K_1=40, K_2=20)$.
- Compute mle for $\theta$.
I assumed that $\theta$ is the probability of Bernoulli trails. But how to compute mle and likelihood?
If I understand correctly, each plane has a probability $\theta$ to be hit when passing through each line of defense.
For the first line, $40$ planes were hit, probability is $\binom{100}{40}\theta^{40}(1-\theta)^{60}$.
For the second line, of the $60$ remaining planes, $20$ were hit, probability is $\binom{60}{20}\theta^{20}(1-\theta)^{40}$.
So the likelihood is $$P_\theta(K_1=40,K_2=20) = \binom{100}{40}\theta^{40}(1-\theta)^{60}.\binom{60}{20}\theta^{20}(1-\theta)^{40} = \frac{100!}{(40!)^2 20!}\theta^{60}(1-\theta)^{100}$$ Taking the derivative, you find $$P_\theta(K_1=40,K_2=20) = \frac{100!}{(40!)^2 20!}\theta^{59}(1-\theta)^{99}\left[60-160\theta\right]$$ so the maximum likelihood is $\theta=\dfrac{60}{160}=\dfrac38$.