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Solution to a bag contains 4 balls Two balls are drawn at random and are found to be white What is the probability that all balls are white?
Two balls are certainly white. So the remaining balls may be one of the following
{WW, WN, NW, NN}
where W means a white ball and N means non-white
ball
Let the events be:
E- all four balls are white(WW)
F- 3 balls are white(WN,
NW)
G- only two balls are white(NN)
Now, P(E) =
P(G) =
1/4, and P(F) =
1/2
Let A be the event that the two balls drawn are white
P(A|E)
=
4C2/4C2 =
1
P(A|F)
=
3C2/4C2 =
1/2
P(A|G)
=
2C2/4C2 =
1/6
Now By Bayes' Theorem,
required probability =
P(E|A)
=
(1*1/4) / ( 1*1/4 + 1/2*1/2 + 1/6*1/4 )
=
(1/4) / ( 13/24 )
=
6/13
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2nd opinion
Let's say we have 3 boxes with 4 balls each.
Box A has 4 white balls.
Box B has 3 white balls.
Box C has 2 white balls.
The probability of drawing 2 W balls from;
Box A, P(2W│A) =
(4/4)∙(3/3) =
1
Box B, P(2W│B) =
(3/4)∙(2/3) =
1/2
Box C, P(2W│C) =
(2/4)∙(1/3) =
1/6
Say the probability of picking any of the 3 boxes is the same, we have;
P(A) =
1/3
P(B) =
1/3
P(C) =
1/3
Question is, given the event of drawing 2 W balls from a box taken blindly
from the 3 choices, what is the probability that the balls came from box A,
P(A│2W).
Recurring to Bayes Theorem:
P(A│2W) =
[P(A)P(2W│A)]/[P(A)P(2W│A)+P(B)P(2W│B)+P(C)P(2W│C)] =
[(1/3)(1)]/[(1/3)(1)+(1/3)(1/2)+(1/3)(1/6)] =
6/10 =
0.60 =
60%
P(A│2W) =
0.60 =
60%
What else can I help you with?
What is the probability of having 2 red balls when chosen in random from a container has 2 red balls 4 green balls and 2 yellow balls?
1/4 chance
If an urn contains 6 yellow balls 3 red balls and 5 blue balls If 2 balls are drawn without replacement what is the probability of drawing 2 yellow balls?
The probability is (6/14)*(5/13) = 30/182 = 0.1648 approx.
If A box contains three black balls and four gold balls Two balls are randomly drawn in succession from the box If there is no replacement what is the probability that both balls are black?
(3/7)*(2/7)=(6/49) You have a 6 out of 49 probability.
What is the probability of choosing 3 red balls and 2 black balls out of a bag which contains 5 red balls 2 green balls and 4 black balls?
If you draw enough balls, without replacement, the probability is 1.The answer depends onhow many balls are drawn, andwhether or not they are replaced.Unfortunately, your question gives no information on these matters.
A bag contains 3 red balls and 2 blue balls a ball is taken at random from the bag and then put back what is the probability that both balls are red?
3c2+2c2+3c1 2c1 --------------- 7c2. I have done it this way. 3C2 means that the both balls would be green 2C2 both would be red and 3C1 and 2C1 because both the balls can be red and green.
8 identical balls are placed at random in three bags. what is the probability that the first bag will contain 3 balls?
The probability that it contains exactly 3 balls is 6/45 = 0.133... recurring.
When jar contains 6 blue balls 8 red balls and 6 yellow balls 2 balls selected without replacement.what is probability that first ball is red and second ball is blue?
If the balls are selected at random, then the probability is 12/95.
A bucket contains 50 lottery balls numbered 1-50. One is drawn at random. What is the probability of the number being 2- digit number?
40/50
What is the probability of having 2 red balls when chosen in random from a container has 2 red balls 4 green balls and 2 yellow balls?
1/4 chance
If you have 4 green balls 2 red balls and 1 blue what is probability of getting a green?
It is 1/7 if a ball is selected at random.
One ball is drawn at random from a bag containing 17 red balls and 4 white balls what is the probability that the ball is red?
17 out of 21
You are going to take a ball at random from a bag. The probability it will be red is two-fifths. There are eight red balls in the bag. How many of the balls are not red?
12
If an urn contains 6 yellow balls 3 red balls and 5 blue balls If 2 balls are drawn without replacement what is the probability of drawing 2 yellow balls?
The probability is (6/14)*(5/13) = 30/182 = 0.1648 approx.
If A box contains three black balls and four gold balls Two balls are randomly drawn in succession from the box If there is no replacement what is the probability that both balls are black?
(3/7)*(2/7)=(6/49) You have a 6 out of 49 probability.
When an urn contains 12 red balls 153 blue balls 2 black balls 47 white balls and no green balls What is the probability you do not draw any ball at all?
2
A box contains four balls two red and two yellow balls we extract two balswhat is the probability of extracting two red balls?
It is 1/6.
What is the probability of choosing 3 red balls and 2 black balls out of a bag which contains 5 red balls 2 green balls and 4 black balls?
If you draw enough balls, without replacement, the probability is 1.The answer depends onhow many balls are drawn, andwhether or not they are replaced.Unfortunately, your question gives no information on these matters.
