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A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft 22 Jan 2021, 05:30
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E
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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 \(P_B\) is the probability, after the first two selections have been made, that the third selection is an apple. 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

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E
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A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft 22 Jan 2021, 06:27
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Bunuel
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
Show more

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.
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A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft 02 Feb 2021, 05:22
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can someone explain why Pb>Pa
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A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft 02 Feb 2021, 11:02
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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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A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft 02 Feb 2021, 11:34
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can someone explain why Pb>Pa
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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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A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft 03 Feb 2021, 10:31
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fall2021
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can someone explain why Pb>Pa

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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Bunuel
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
Show more
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.
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A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft 03 Feb 2021, 15:36
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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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A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft 04 Feb 2021, 03:43
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fall2021
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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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.
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A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft 09 Feb 2021, 15:08
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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.
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A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft 03 Dec 2023, 04:29
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Bunuel
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

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.
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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
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A box has 4 apples and 3 oranges. Robert picks fruits randomly one aft 24 Aug 2024, 00:32
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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 [apple-apple + apple-orange + orange-orange). My reasoning was that summing these probabilities would give me the overall likelihood of picking an apple as the third fruit

Turns out I was wrong because the OA is E, which I understand that I should compare 4/7 vs. 2/5, vs. 3/5, vs. 4/5 (therefore i. and iii. is possible)

Questions:
1. Why shouldn't we add the probabilities 2/5 + 3/5 + 4/5 to know P(B)?
2. How can we differentiate whether P(B) is a 3 single events that should be compared with P(A) and P(B)? I want to be able to avoid this mistake further

Thanks!

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