A toy manufacturing company tests the quality of the toys it manufactures. On a daily basis, 20 toys are taken at random for testing and to check that 95% meet the toy specification. If more than 2 of the 20 toys are of an unacceptable standard then the production must be temporarily stopped.
Calculate the probability if the inspectors pass the day's toy production as acceptable if 75% of the day's toys meet the specification.
$n = 20$
$Success(P) = 1 - 0.75 = 0.25$
$Failure(Q) = 0.75$
$x = (20*0.25) = 5$
$= (^{20}C_5) (0.25)^5 (0.75)^{20-5}$
$= 0.202$
Is this correct? If my answer is incorrect, where have I gone wrong? Thanks.
I think you've made a couple mistakes. For one, I think you should consider "success" to be "the toy meets the specification," so $P(\text{success})=0.75$ and $P(\text{failure})=0.25$.
Now you wrote "if more than $2$ of the 20 toys are of an unacceptable standard then the production must be temporarily stopped." You should check that this isn't supposed to say "if $2$ or more of the $20$ toys are..." because if we go based on the statement that $95\%$ of the toys have to be acceptable, then at most one toy can fail to meet the specification.
Next you compute $x=20\cdot 0.25$, i.e., $n \cdot p$ where $n$ is the number of trials and $p$ is the probability of "success" (as you have defined it). This tells you the expected number of "successes" (i.e., we expect $5$ of the $20$ toys to not meet specifications). This calculation is not relevant to answering the question, though.
Rather, if $X$ denotes the number of toys which meet specifications, we want to know the probability that $X$ is large enough that $95\%$ of the toys meet specifications. That is, we want $P(X \geq 19)$. Since $X$ can only take on the values $0,1,\dots,20$, we have $P(X \geq 19)=P(X=19)+P(X=20)$. Now using the binomial formula, for any $0 \leq k \leq 20$ we have $$P(X=k)= (^{20}C_k)(0.75)^k (0.25)^{20-k}.$$ I think you can finish it from there.