Question:
Eggs are sold in boxes of six and it is likely that 1% of the eggs will be broken when they are unpacked. Find:
(i) the probability that a box contains no broken eggs (ii) the probability that a box contains no more than one broken egg
I buy four boxes. Find:
(iii) the probability that I get no broken eggs (iv) the probability that I have at least two broken eggs
I know the answer to the first 3 questions:
(i) 0.9415 (ii) 0.9985 (iii) 0.7857
But the fourth one is killing me!
Could you show the steps of the answers as well? Thank you so much!!!
With $p=1\%=0.01$ we have $$ P(\left\{\text{$k$ of $n$ eggs broken}\right\})=\binom{n}{k}p^k(1-p)^{n-k} $$ Knowing this, you can evaluate $$\begin{align} P(\{\text{$0$ of $6$ eggs broken}\})&=\binom{6}{0}p^0(1-p)^{6-0}=(1-p)^6\tag{$i$}\\ P(\{\text{$0$ or $1$ of $6$ eggs broken}\})&=P(\{\text{$0$ of $6$ eggs broken}\})+P(\{\text{$1$ of $6$ eggs broken}\})\\&=\binom{6}{0}p^0(1-p)^{6-0}+\binom{6}{1}p^1(1-p)^{6-1}\\&=(1-p)^6+6p(1-p)^5\tag{ii}\\ P(\{\text{$0$ of $24$ eggs broken}\})&=\binom{24}{0}p^0(1-p)^{24-0}=(1-p)^{24}\tag{$iii$}\\ P(\{\text{more than $1$ of $24$ eggs broken}\})&=1-P(\{\text{$0$ or $1$ of $24$ eggs broken}\})\\&=1-\binom{24}{0}p^0(1-p)^{24-0}-\binom{24}{1}p^1(1-p)^{24-1}\\&=1-(1-p)^{24}-24p(1-p)^{23}\tag{iv} \end{align}$$