The probability that a randomly selected box of a certain type of cereal has a particular prize is 0.2. Suppose you purchase box after box until you obtained two of these prizes. What is the probability that four boxes are purchased?
My approach was to use negative binomial distribution:
Let $X$ be the number of boxes purchased that don't have a prize until you find two prizes.
P(4 boxes purchase) = P(2 boxes w/o a prize)
=$$\binom{x+r-1}{r-1}(p)^r(1-p)^x$$ = $$\binom{2+2-1}{2-1}(0.8)^2(1-0.2)^2$$ = $$\binom 3 1(0.2)^2(0.8)^2$$ = $$.0764$$
I wanted to know if this is correct approach to solving this kind of probability question?
Best Answer
I would note that there are three paths to this result
--++
-+-+
+--+
each with a probability of $0.8^2\cdot0.2^2=0.0256$. Thus, the total probability would be $3\cdot0.0256=0.0768$
This is a slightly different approach getting to the same result.