A Bag contain 5 black 6 white and 7 red balls four balls are drawn at random from it.If x denotes the number of white balls,then findE(x)
Answer (1)
25th Jan, 2021
Dear student,
Total number of balls in bag = 18 out of which 6 are white balls and other balls count to a total of 12.
It is given that X is the number of the white balls drawn. And the white balls are drawn four in number.
Now, since nothing is given so, we can assume that ball are drawn without replacement.
So, expectation E(X) = summation (XP(X))
So, let P(X=0) = all drawn balls are non-white. So X = 0 so, E ( X = 0) = 0
Now, probability of drawing one white ball, i.e P(X=1) = 4 {(6/18) (12/17) (11/16) (10/15)} = 0.4314
Since we are not told any order so there can be four cases that a White ball is drawn in first attempt or in second attempt or in third attempt or in fourth attempt or you cam saya 4C1 cases.
Now, P( X=2 ) = 6 {(6/18) (5/17) (12/16) (11/15)} = 0.3235
P(X=3) = 4 {(6/18) (5/17) (4/16) (12/15)} = 0.0784
And, P(X=4) = {(6/18) (5/17) (4/16) (3/15)} = 0.0049
So, now E(X) = (1 P(X=1)) + (2 P(X=2)) + (3 P(X=3)) + (4 P(X=4))
So, after calculating the values we get E(X) = 1.3332
Note that expectation is not the probability, so it's value can be also greater than 1.
Total number of balls in bag = 18 out of which 6 are white balls and other balls count to a total of 12.
It is given that X is the number of the white balls drawn. And the white balls are drawn four in number.
Now, since nothing is given so, we can assume that ball are drawn without replacement.
So, expectation E(X) = summation (XP(X))
So, let P(X=0) = all drawn balls are non-white. So X = 0 so, E ( X = 0) = 0
Now, probability of drawing one white ball, i.e P(X=1) = 4 {(6/18) (12/17) (11/16) (10/15)} = 0.4314
Since we are not told any order so there can be four cases that a White ball is drawn in first attempt or in second attempt or in third attempt or in fourth attempt or you cam saya 4C1 cases.
Now, P( X=2 ) = 6 {(6/18) (5/17) (12/16) (11/15)} = 0.3235
P(X=3) = 4 {(6/18) (5/17) (4/16) (12/15)} = 0.0784
And, P(X=4) = {(6/18) (5/17) (4/16) (3/15)} = 0.0049
So, now E(X) = (1 P(X=1)) + (2 P(X=2)) + (3 P(X=3)) + (4 P(X=4))
So, after calculating the values we get E(X) = 1.3332
Note that expectation is not the probability, so it's value can be also greater than 1.
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