Find the probability of having $4$ or more girls in a family of $6$ children.
Find also the probability that among $5$ families, each with $6$ children, at least $3$ of the families have $4$ or more girls.
An urn contains $4$ white and $3$ black balls. Balls are randomly selected, one at a time, until a black one is obtained. If we assume that each selected ball is replaced before the next one is drawn, what is the probability that at least $4$ draws are needed?
Suppose that $25%$ of the items taken from a production line are defective. If the items taken from the line production are checked until 10 defective items are found. What is the probability that $15$ items are examined?
For question $1$, is the required probability $3/64$? I just find it out by the calculating the sum of $(0.5)^4 x (0.5)^2 + (0.5)^5 x (0.5) + (0.5)^6$
For question $2$, is the required probability $9.84 x 10^-5$? The method I adopted is similar to the above one.
For question $3$, is that $1 - (3/7 + 4/7 x 3/7 + 4/7 x 4/7 x 3/7 ) = 64/343 $?
Lastly, for question $4$, how do I find the required probability? Please advise me.
(1) also needs to account for the order of births of the children. clearly, probability of having 0,1,...,or 6 girls should be 1 but by your logic you get $$ \sum_{k=0}^6 0.5^k 0.5^{6-k} = 7 \cdot 0.5^6 = 7/64 \neq 1 $$
(2) indeed the method should be similar - need to account for the order also
(3) looks correct to me
(4) look at the Negative Binomial distribution