GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
A bag contains 1 red chip and 9 white chips only. If the chips are ran
[#permalink]
23 Oct 2021, 07:52
23 Oct 2021, 07:52
1
00:00
Question Stats:
100% (01:39) correct
0% (00:00) wrong based on 3 sessions
Hide Show timer Statistics
A bag contains 1 red chip and 9 white chips only. If the chips are randomly selected one at a time without replacement, what is the probability that the last chip selected is red?
A) \(\frac{1}{10^{10}}\)
B) \(\frac{9!}{10^{10}}\)
C) \(\frac{10!}{10^{10}}\)
D) \(\frac{9^9}{10^{10}}\)
E) \(\frac{1}{10}\)
A) \(\frac{1}{10^{10}}\)
B) \(\frac{9!}{10^{10}}\)
C) \(\frac{10!}{10^{10}}\)
D) \(\frac{9^9}{10^{10}}\)
E) \(\frac{1}{10}\)
A bag contains 1 red chip and 9 white chips only. If the chips are ran
[#permalink]
23 Oct 2021, 23:30
23 Oct 2021, 23:30
1
It has to be an arrangement problem. Since the without replacement concept is mentioned in the question.
Since it is given that the last one to be picked has to be the RED chip: "what is the probability that the last chip selected is red?", and as we have only one RED chip, that is an "occurring for sure" probability. So its probability will be "1".
The remaining white chip's probability in the 9 picks will be:
1st Pick: 9(Any white chip can be picked from all the 9)/10(The total chips in the bag) * 2nd Pick:(9-1=8(now one chip is taken out, "and shouldn't be replaced", we are left with only 8 chips)/(10-1=9(Since it's without replacement, now only 9 chips are left))) * 3rd Pick: 7/8 (the same procedure of picks follows) * 6/7 * 5/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1(the "definite case of RED chip being picked in the last").
Now, after all the alternate numerator and denominator gets canceled which will look like:
9/10 * 8/9 * 7/8 * 6/7 * 5/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1:
We'll be left with 1/10. So, It's answer choice E.
Since it is given that the last one to be picked has to be the RED chip: "what is the probability that the last chip selected is red?", and as we have only one RED chip, that is an "occurring for sure" probability. So its probability will be "1".
The remaining white chip's probability in the 9 picks will be:
1st Pick: 9(Any white chip can be picked from all the 9)/10(The total chips in the bag) * 2nd Pick:(9-1=8(now one chip is taken out, "and shouldn't be replaced", we are left with only 8 chips)/(10-1=9(Since it's without replacement, now only 9 chips are left))) * 3rd Pick: 7/8 (the same procedure of picks follows) * 6/7 * 5/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1(the "definite case of RED chip being picked in the last").
Now, after all the alternate numerator and denominator gets canceled which will look like:
We'll be left with 1/10. So, It's answer choice E.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
A bag contains 1 red chip and 9 white chips only. If the chips are ran
[#permalink]
24 Oct 2021, 05:12
24 Oct 2021, 05:12
2
GreenlightTestPrep wrote:
A bag contains 1 red chip and 9 white chips only. If the chips are randomly selected one at a time without replacement, what is the probability that the last chip selected is red?
A) \(\frac{1}{10^{10}}\)
B) \(\frac{9!}{10^{10}}\)
C) \(\frac{10!}{10^{10}}\)
D) \(\frac{9^9}{10^{10}}\)
E) \(\frac{1}{10}\)
A) \(\frac{1}{10^{10}}\)
B) \(\frac{9!}{10^{10}}\)
C) \(\frac{10!}{10^{10}}\)
D) \(\frac{9^9}{10^{10}}\)
E) \(\frac{1}{10}\)
I created this question to highlight the myth that the probability of randomly selecting a specific object decreases as more objects are selected.
In actuality, P(1st chip is red) = P(2nd chip is red) = P(3rd chip is red) = . . .. = P(9th chip is red) = P(10th chip is red) = \(\frac{1}{10}\)
Answer: E
We can also perform the calculation as follows:
P(10th chip is red) = P(1st is white AND 2nd is white AND 3rd is white AND 4th is white AND.... 8th is white AND 9th is white AND 10th is RED)
= P(1st is white) x P(2nd is white) x P(3rd is white) x P(4th is white) x. . . . x P(8th is white) x P(9th is white) x P(10th is RED)
= 9/10 x 8/9 x 7/8 x 6/7 x 5/6 x 4/5 x 3/4 x 2/3 x 1/2 x 1/1
A lot of numerators and denominators cancel out to get.....
= 1/10
