We have a box of balls with color's green and blue. Percentage of green ball in the box is 55%. If we were to choose 12 balls at random from the box then what is the probability that exactly 5 of them will be green? Also what is the probability that at most eight of them will be green? what is the probability that at least seven of them will be green? what is the probability that at least 5 of them , not more than 9 will be green?
what I tried
For first question what is the probability that exactly 5 of them will be green?
(0.55**5) = 0.05
But this looks like is wrong
Assuming that there are a sufficiently large amount of balls so that draws can be considered independent i.e. $N>12\cdot 10$, then we can say that each draw has a $.55$ chance of being green.
From there using binomial coefficients, the probability that there are exactly $k$ green balls is $$P(k)=\binom{12}{k}(.55)^k(1-.55)^{12-k}$$
Hence, the answers to your four questions are respectively $$P(5)$$ $$\sum_{k=0}^8 P(k)$$ $$\sum_{k=7}^{12} P(k)$$ $$\sum_{k=5}^9 P(k)$$