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Tabatha decorates the walls of her room by pasting 80 balloons, which include 20 red balloons. She randomly selects one balloon at a time and then pastes it on the wall. What is the probability that the third balloon she selects is a red balloon?
(A) 1/80
(B) 1/20
(C) 1/4
(D) 1/2
(E) None of the above
(A) 1/80
(B) 1/20
(C) 1/4
(D) 1/2
(E) None of the above
Praveena_10
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Probability of Selecting a red balloon= 20/80=1/4
Probability of selecting other colour balloon=60/80=3/4
Let X be other colour
There are 4 cases we can get for the above situation
RRR,RXR,XRR,XXR
Probability of selecting R in third balloon=1/4*1/4*1/4+1/4*3/4*1/4+3/4*1/4*1/4+3/4*3/4*1/4=16/64=1/4.
Probability of selecting other colour balloon=60/80=3/4
Let X be other colour
There are 4 cases we can get for the above situation
RRR,RXR,XRR,XXR
Probability of selecting R in third balloon=1/4*1/4*1/4+1/4*3/4*1/4+3/4*1/4*1/4+3/4*3/4*1/4=16/64=1/4.
sarcastic__atom
The point is that we don't know what the first two balloons colors were, so without that information, nothing changes when calculating the probability for the third balloon.
Consider this: there are 8 cards, 5 spades, and 3 hearts. What is the probability that the first card you pick will be a spade? Clearly, it’s 5/8.
Now, suppose I randomly remove three cards without telling you which cards were thrown away. What is the probability now that you pick a spade out of the 5 remaining cards? Even though the sample has been reduced, the probability still remains 5/8. Why should it change?
Another scenario: using the same 8 cards (5 spades and 3 hearts), imagine you are allowed to pick one card, but only from a randomly selected subset of 5 cards. What is the probability of picking a spade? Again, it remains 5/8.
In all these cases, the probability does not change. If it were to increase or decrease, it would imply that the probability of picking a heart would inversely change. But there is no reason for this to happen, as the process remains random, and we lack information about the removed or previously picked cards.
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Posts: 3
If she selects and paste it on the wall then how come the probability still remains 1/4 it should be 19/79 right ?
Yes, we don't know what the first 2 picks were (red or non red), but doesn't that mean we should account for all the possible cases for 1st and 2nd and accordingly account for the 3rd one? Since, there is NO REPLACEMENT also in this question, balloon gets pasted on the wall, so there is a difference in remaining number of total balls and remaining number of red balls-which will differ the likelihood or probability. Like in the following questions also, we did the same thing, probability of what we were supposed to calculate was dependent upon what the previous scenarios were. Attaching the links to the similar questions I'm talking about-
1) https://gmatclub.com/forum/there-are-10 ... 98979.html
2) https://gmatclub.com/forum/there-are-7- ... 57431.html
I'm missing some logical gap in understanding about the shorter method i.e 1/4 probability for all the picks in our current case. Please help me in understanding what am i missing.
Thanks
1) https://gmatclub.com/forum/there-are-10 ... 98979.html
2) https://gmatclub.com/forum/there-are-7- ... 57431.html
I'm missing some logical gap in understanding about the shorter method i.e 1/4 probability for all the picks in our current case. Please help me in understanding what am i missing.
Thanks
Bunuel
durga1024
Read more carefully. The question asks only for the probability that the third balloon is red, not about any sequence before it. Since every balloon is equally likely to be in any position, the probability that the third balloon is red is simply the overall proportion of red balloons: 20 out of 80, or 1/4. For better understanding, practice the similar questions linked there.
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