a) What's the probability of selecting a red apple when choosing at random?
This one is easy enough for me it's just $P(R) = \cfrac{2}{6}$ where R is the event we pick a red apple
b) What's the probability that, if one apple is randomly chosen per day and then eaten and not replaced, red apples are chosen on the first two days and green apples are chosen on the last four days.
I'm a little confused on this one, please let me know if my methodlogy/reasoning is correct.
There are $\cfrac{6!}{4!2!}=15$ ways/permutations to pick and eat the apples when we don't care about any constraints
Red apples being eaten on the first two days and green apples being eaten the last four days is ONE specific permutation of the $15$ permutations. So I believe that the answer is the Probability $= \cfrac{1}{15}$ Do I have b right? Please help me note any flaws in my thinking
Best Answer
The order in which we need to pick the apples is $\mathrm{RRGGGG}$. The first time we pick apples, probability of picking $\mathrm{R}$ is $\frac 26 = \frac 13$. After that, we're left with 1 red apple and 4 green apples. The proabbility of picking a red apple on the second day is then $\frac 15$. After this, since all the apples left are green apples, the probability of picking a green apple is $1$ on all following days.
The final probability is then $P = \frac {1}{3} \times \frac{1}{5} = \frac{1}{15}$