I'm building onto my previous question, but I will include it also in here. I'm again struggling with the probability questions we've been given.
Six balls numbered 1,2,2,3,3,3 are placed in a bag. Balls are taken one at a time from the bag at random and the number noted. Throughout the question a ball is always replaced before the next ball is taken.
The questions are following:
A) Find the least number of balls that must be taken from the bag for the probability of taking out at least one ball numbered 2 to be greater than 0.95.
B) Another bag also contains balls numbered 1,2 or 3. Eight balls are taken from the bad at random. It is calculated that the expected number of balls numbered 1 is 4.8, and the variance of the number of balls numbered 2 is 1.5. Find the least possible number of balls numbered 3 in this bag.
I would really appreciate your help or at least hints..
Best Answer
For (a), let $A$ be the event that at least one $2$ is drawn in $n$ draws. Then $$ P(A)=1-P(A^c)=1-\left(\frac{4}{6}\right)^n $$ We want $P(A)\geq 0.95$ which you can solve for $n$.
For the second question let $X, Y, Z$ be the number of balls numbered $1$, $2$, $3$ drawn from the $8$ draws respectively. Then $X, Y, Z$ are binomially distributed (since the balls are replaced before being drawn). In particular, it is given that $$ EX=4.8=8p_1 $$ $$ \text{Var}(Y)=1.5=8p_2(1-p_2) $$ where $p_i$ is the probability of drawing a ball numbered $i$ on a single trial for $i=1,2,3$. You can solve for $p_1, p_2$ and hence $p_3$ which will yield the answer to (b.)