GDT
Bunuel
11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. 7/8
B. 3/4
C. 2/3
D. 5/8
E. 3/8
Basically we need to find the probability that the seventh marble drawn is red (so not blue).
Now, the initial probability of drawing red marble is 5/8. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is 5/8.
The same for blue marble: the probability of drawing blue marble is 3/8, the probability that for instance the 8th marble drawn is blue is still 3/8. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).
Answer: D.
MentorTutoringIn this question we have to draw w/o replacement, so wouldn't the probability of drawing balls be different - 1st ball red 5/8, 2nd ball red 4/7...
Pls explain how to do this
Thanks in advance!
I know it may be counterintuitive,
GDT, but the answer should be (D). Picture eight slots in the following manner:
___ ___ ___ ___ ___ ___ ___ ___
Now, if I told you I
had drawn all eight marbles and asked specifically about any given slot, would you not agree that the answer would be 5/8 for any particular draw that was NOT blue? That is, all you can say here is that for any given slot, there would be only two options, red or blue, and there happen to be 5 red and 3 blue marbles. If you
knew what had been selected first, second, third, and so on, the probability would change on a sliding scale, as you have indicated. A specific designation per selection is the more typical variant of this concept.
This problem comes to mind, for example. In the slots above, however, there is no way to distinguish one selection from another without more information.
Out of curiosity, what answer would you propose from the choices above if you did attempt to solve the problem in the manner you have outlined?
- Andrew
I was trying to make cases-1. when 2 left in the end would be red, 2.when out of 2 left in end one would be red and other blue