Paula moves to an area with different telephone exchange. Telephone numbers start with 753 and all combinations of the remaining 4 digits is equally likely. What is the probability that the last 4 digits in Paula's new telephone number are odd?
I tried to do this, but it doesn't give me the answer:
I said n = 4 I honestly don't know what x is anymore..
p = 0.5 since there are 5 numbers that are odd through 0-9
q= 0.5 since there are 5 numbers that are odd through 0-9, maybe i got this part wrong?
Yes, I think you have it now. You have noticed that $X\sim$Binomial, or $X$ follows a binomial distribution. A known probability function for this is $$P(X=x)=\binom{n}{x}p^xq^{n-x}.$$
$n=4$, for the four digits or "trials."
$p=0.5,$ because probability of getting an odd number is $\frac{5}{10}$ for each digit.
$q=1-p=0.5,$ since this is the probability of getting "not odd."
Then we try to find $P(X=4)$, which is the probability that all four digits are odd.