So I encountered this problem while I was studying for exam. However, I cannot wrap my head around the solution that the answer key provided.

The problem goes like this:

Bob watches cars that pass by his house. Bob assumes that each successive car that goes by has a $\frac{1}{4}$ chance of being a Japanese car and $\frac{3}{4}$ chance of being an American car. If his assumption is right, what is the probability that Bob will see at least two Japanese cars pass by before the $3^{rd}$ American car passes by.

From my understanding, the perfect case would be only five cars intotal$(2\ Japanese\ cars\ and \ 3\ American\ cars)$ pass by his house and he sees the $3^{rd}$ American car the last. Such as, $JJAA\ or\ JAJA$, and then followed by the $3^{rd}A$, the probability of this case would be $6\cdot(\frac{1}{4})^2\cdot(\frac{3}{4})^3$. I realized that this may not be the right approach because it could be the case like $JJJJAAA$ or even more $A's$ before three $D's$. I also tried to approach the problem by using complement of $A\geq2$ with $A=0\ and \ A=1$. But all of this is still under the assumption of only five cars intotal pass by. I'm stuck now....

Any help would be appreciated.

  • $\begingroup$ Try solving the complementary problem: probability of seeing less than two J's before the third A. $\endgroup$ – Moss Murderer Jan 21 at 7:32

This is simply $1-$ P($AAA$, $JAAA,AJAA,AAJA$). Because, in all other situations, you get two Japanese cars before the 3rd American car. These have $(1-p)^3,p^2(1-p)^3,p^2(1-p)^3,p^2(1-p)^3$ probabilities respectively. Summing them yields, $(1+3p)(1-p)^3=189/256$, subtracting from $1$ yields $67/256$.


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