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22 Nov 2021, 09:15
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A
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A bag contains 6 balls of one or more colors. A ball is picked and is found to be red. What's the probability that the bag initially had exactly 6 red balls?
A. 2/7
B. 1/5
C. 3/5
D. 4/5
E. 1/7
A. 2/7
B. 1/5
C. 3/5
D. 4/5
E. 1/7
22 Nov 2021, 11:34
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Bunuel
Breaking Down the Info:
If there was only 1 red, there is a \(\frac{1}{6}\) chance of getting the red ball first. We would have \(\frac{2}{6} =\frac{1}{3}\) chance of getting the red ball if there was 2 red, etc.
If we assume 1 ball to 6 balls each have an equal chance of happening (we don't include the 0 red ball case because we can't pull out a red ball if there are 0 red balls), then there is a \(\frac{(\frac{1}{6} + \frac{2}{6} + \frac{3}{6} +...+1)}{6} = \)\(\frac{21}{36}\) chance of pulling a red ball first in general.
There is a \(\frac{1}{6}\) chance for 6 red balls among the 6 cases.
Then the probability for 6 red balls given 1st ball is red from Bayes theorem is :
P(6 Red Balls | 1st Ball red) = P(1st Ball Red | 6 Red Balls) * P(6 Red Balls) / P(1st Ball Red) = \(100\% * \frac{1}{6} * \frac{36}{21} = 6/21 = \frac{2}{7}\).
Answer: A
22 Nov 2021, 11:43
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1 red,5 different colors: 1/6 of picking red
2 red,4 " 2/6
.
.
6 red, 0 ". 1
Add probabilities;
(1+2+3+4+5+6)/6= 21/6
6 red probability; 6/6
Probability of 6 red balls given red pick:
(6/6)/(21/6) = 6/21= 2/7
Posted from my mobile device
2 red,4 " 2/6
.
.
6 red, 0 ". 1
Add probabilities;
(1+2+3+4+5+6)/6= 21/6
6 red probability; 6/6
Probability of 6 red balls given red pick:
(6/6)/(21/6) = 6/21= 2/7
Posted from my mobile device
