a bag contain 3 red & 7 white marbles. a marbel is drawn from the bag & a marbel of the colour is then put into the bag a second marbel is drawn from the bag of both marvel war of the same colour. wat is the probability that they were both white ??
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3 red balls 7 white balls.
P(Red) = 3/10 P(White) = 7/10
As the first marble drawn is put back in the bag, probabilities remain same for the second marble drawn from the bag. Both events of drawing of marbles are independent of each other.
Probability that both marbles are Red
= P(first marble = Red AND 2nd marble = Red)
= 3/10 * 3/10 = 0.09
Probability that both are white
= P(1st marble = White AND 2nd marble = White)
= 7 /10 * 7/10 = 0.49
Probability that both marbles drawn are of same color
= 0.09 + 0.49 = 0.58
Given that both marbles are of the same color. Then probability that both happened to be white = 0.49 / 0.58 = 0.845.
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We get this by applying the conditional probability theorem.
P (B | A) = P( B Π A ) / P(A)
Here event B = both marbles are White.
A = both marbles are of same color.
B Π A = both marbles are White = B
P(Red) = 3/10 P(White) = 7/10
As the first marble drawn is put back in the bag, probabilities remain same for the second marble drawn from the bag. Both events of drawing of marbles are independent of each other.
Probability that both marbles are Red
= P(first marble = Red AND 2nd marble = Red)
= 3/10 * 3/10 = 0.09
Probability that both are white
= P(1st marble = White AND 2nd marble = White)
= 7 /10 * 7/10 = 0.49
Probability that both marbles drawn are of same color
= 0.09 + 0.49 = 0.58
Given that both marbles are of the same color. Then probability that both happened to be white = 0.49 / 0.58 = 0.845.
====
We get this by applying the conditional probability theorem.
P (B | A) = P( B Π A ) / P(A)
Here event B = both marbles are White.
A = both marbles are of same color.
B Π A = both marbles are White = B
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