fall2021 wrote:
sthahvi wrote:
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
Posted from my mobile deviceBunuel wrote:
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
Hey
fall2021Don'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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