A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft

Question

A box has 4 apples and 3 oranges. Robert picks fruits randomly one after the other without replacement. Probability of the 1st fruit as apple is \(P_A\) and that of the 3rd fruit is apple is \(P_B\). Which of the following could true?

I. \(P_A>P_B \)

II. \(P_A=P_B\)

III. \(P_A<P_B\)

A. I only
B. II only
C. III only
D. I and II only
E. I and III only

I.  P_A>P_B

II.  P_A=P_B

III.  P_A<P_B

A. I only
B. II only
C. III only
D. I and II only
E. I and III only

vijaykhot · Answer

P_A = 4/7
P_B = 3/7 * 2/6 * 4/5 = 4/35 (If the first two fruits are Oranges and the third is an apple)
P_B = 4/7 * 3/6 * 2/5 = 4/35 (If the first two fruits are Apples and the third is an apple)
P_B = 3/7 * 4/6 * 3/5 = 6/35 (If the first fruit is an orange, the second an apple and the third is an apple)
P_B = 4/7 * 3/6 * 3/5 = 6/35 (If the first fruit is an apple, the second an orange and the third is an apple)

For all cases P_A > P_B,
Hence, A.

sthahvi · Answer

can someone explain why Pb>Pa

lordwells · Answer

Hi Bunuel...can you kindly explain this?...IMO there is a line of reasoning potentially pointing to B as the correct answer, but I suspect it is wrong...Thanks

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fall2021 · Answer

Hi, the answer doesn't imply Pb is definitely greater than Pa. The question asks which of these "could" be true.

In the case where first 2 fruits picked are oranges, Pb = 4/5 which is greater than Pa (= 4/7)

Similarly when first 2 picks are apples then Pb = 2/5 which is less than Pa

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unraveled · Answer

Hey fall2021 Don't you think the case where first two picks are oranges is not valid at all as we are dealing with apples only. So, 1st and 3rd will always be apples. I think E is answer because 1st( P_A = \frac{4}{7} ) will always be apple and then we have following cases: 1. If 2nd is apple then P_B = \frac{2}{5} \implies P_A > P_B ( \frac{4}{7}>\frac{2}{5} ) 2. If 2nd is not apple then P_B = \frac{3}{5} \implies P_A < P_B ( \frac{4}{7}<\frac{3}{5} ) That's why both I and III can be true. Answer is E then.

fall2021 · Answer

Hi, Pa and Pb are not interlinked. So, it's not necessary that while calculating Pb first pick has to be Apple.

The answer works out to be E either way but to calculate highest possible Pb we should consider the case where first 2 picks are oranges.

The question doesn't say Pb is probability of picking Apple in 3rd turn given an apple is picked first. It just defines Pa and Pb as 2 different probabilities.

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unraveled · Answer

Yes, we can say that and I'm not saying they are interlinked.
But my point was based on "Probability of the  1st fruit as apple  is  P_A  and that of the  3rd fruit is apple  is  P_B  ".

The condition "given" is out of question since both are independently mentioned. 
Anyway, the question becomes easier when it specifically mentions  P_A  and  P_B  being probabilities of picking apple only.

IanStewart · Answer

The question doesn't make sense. As I'd interpret it, the third fruit is just as likely to be an apple as the first fruit, if you have no information about the first two selections. So P(A) = P(B) = 4/7. It's just like picking the first card and the third card from a deck of cards. They each have a 1/4 probability of being a diamond.

Since the OA here is E, what the question presumably means to ask is:  if P(A) is the probability the first selection is an apple, and P(B) is the probability,  after the first two selections have been made , that the third selection is an apple, what could be true?  Then P(A) is 4/7, and P(B) is something out of 5, so P(A) cannot be equal to P(B). P(B) can be greater than P(A) if we remove oranges with the first two picks, and otherwise will be less than P(A). But if that's what the question means, it needs to say that -- no mathematician in the universe would interpret this question, as written, in this way.

ruis · Answer

We don´t multiply propobabilities in P(B) because they are not independent events, correct? 
P(B) for all the scenarios described are:
4/5
2/5
3/
3/5
This is to say P(B) is just the probability of the thrid case.
So, you compare the smallest and the largest. 4/7 > 2/5 and 4/7 < 4/5. Hence, answer choice E

heron6 · Answer

I initially believed that the correct approach to compare P(A) and P(B) was to treat P(B) as a combination of different scenarios (which is 2/5 + 3/5 + 4/5, combining 3 scenarios based on first two fruits

A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft

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A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft

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