A bag contains 5 white and 3 black balls, and 4 balls are successively drawn out and not replaced. Find the probability that they are alternatively of different colors.

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Sabyasachi Dey

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it can be in two ways-( white, black, white., black.) or ( black, white, black, white). in first case, the probability is=(5/8)X(3/7)X(4/6)X(2/5)=1/14. IN SECOND CASE, IT IS=(3/8)X(5/7)X(2/6)X(4/5)=1/14. SO THE PROBABILITY IS=(1/14)+1/14)=1/7
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Pursuing B.E from Punjab university.

probability is (5*3*4*2 + 3*5*2*4) / (8*7*6*5)
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Math Teacher

5C2*3C2-2/8C4
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Math Teacher

let me know the the balls are alike or different....because the event for same balls =3c2*1*1 and different for different ball
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Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

Required Probability = P(WBWB) + P(BWBW) = 1/14 + 1/4 = 2/14 = 1 / 7
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Two possible orders=> WBWB or BWBW in both cases the probability is (5*3*4*2)/(8*7*6*5) so counting both the orders the probability is (5*3*4*2)*2/(8*7*6*5) =(1/7)
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Tutor

There are 2 cases:- BWBW & WBWB Case1: probability for BWBW =(3/8)*(5/7)*(2/6)*(4/5) = 1/14 Case 2: WBWB = (5/8)*(3/7)*(4/6)*(2/5) = 1/14 So, total probability = 2/14 = 1/7
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Genuine artist in teaching mathematics in a very very simple yet exciting way

(5/8 * 3/7 * 4/6 * 2/5) + (3/8 * 5/7 * 2/6 * 4/5)
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Teacher

As explained by Sabyasachi above, the required probability is 1/7
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Teaching is my passion!!!!

1/7
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