Probability: 2 ENGINE for rocket a and 4 four rocket b One rocket has two engines and another has four. All engines are identical. Each rocket will achieve its mission if and only if at least half its engines work. Both rockets have the same nonzero probability of achieving its mission. What is the probability that any particular engine works?
I tried $${4 \choose 2}\cdot p^2\cdot(1-p)^2={2 \choose 1}\cdot p\cdot(1-p)$$
Based on these assumptions:


*

*Because $p$ will be the same for any given engine.

*Also if the first rocket achieves its mission and the second also achieves his    then I can write $P(X=1)=P(X=2)$
where $X =$ number of working engine

 A: You missed the fact that the mission is achieved if at least half the engines work, so for the two engine rocket you need to add in the chance that both engines work and for the four engine rocket you need to add in ....
A: The probability that at least $1\over2$ of the engines work for the rocket with $4$ engines is $${\sum_{k=2}^{4}} {4\choose{k}}p^k (1-p)^{4-k}$$
The probability that at least $1\over2$ of the engines work for the rocket with $2$ engines is 
is $${\sum_{k=1}^{2}} {2\choose{k}}p^k (1-p)^{2-k}$$
Setting these equal to each other, we get
$${4\choose{2}}p^2 (1-p)^{4-2}+ {4\choose{3}}p^3 (1-p)^{4-3}+{4\choose{4}}p^4 ={2\choose{1}}p^1 (1-p)^{2-1}+{2\choose{2}}p^2$$
With some algebra you will find that this gives
$$3p^4 -8p^3 +7p^2 -2p=0$$
This has roots of $p=0$, $p=1$ and $p$$=$$2\over3$ but $p$ obviously can't be zero or one so that gives us $p$$=$$2\over3$.
One can use R to check that the following is true:
$${4\choose{2}}{2\over3}^2 \cdot {1\over3}^{2}+ {4\choose{3}}{2\over3}^3 \cdot {1\over3}+{4\choose{4}}{2\over3}^4 ={2\choose{1}}{2\over3} \cdot {1\over3}+{2\choose{2}}{2\over3}^2$$
sum(dbinom(2:4, 4, 2/3))
sum(dbinom(1:2, 2, 2/3))
> sum(dbinom(2:4, 4, 2/3))
[1] 0.8888889
> sum(dbinom(1:2, 2, 2/3))
[1] 0.8888889

