There are 12 white and 12 red balls in a bag. Balls are drawn one by one with replacement from the bag. What is the probability that 7th drawn ball is 4th white ball ?

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To find: The probability that the 7th ball drawn is the 4th white ball. Solution: Probability of getting a white ball per draw is 1/2. Probability of getting a red ball per draw is 1/2. As the balls are replaced after every draw, this probability does not change. Now for a clear understanding...
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To find: The probability that the 7th ball drawn is the 4th white ball. Solution: Probability of getting a white ball per draw is 1/2. Probability of getting a red ball per draw is 1/2. As the balls are replaced after every draw, this probability does not change. Now for a clear understanding let us consider each draw as an event. there are total seven events. _ _ _ _ _ _ _. For the seventh event to be a white ball, the previous six events should contain 3 white balls and 3 red balls in any random arrangement. The number of ways for this is 6C3 (1/2)^3 * (1/2)^3 [6C3 is the number of ways to select 3 places out of 6 places. (1/2)^3 is probability of getting 3 white balls and the second (1/2)^3 is the probability of getting three red balls]. = [(6*5*4)/(3*2*1)] * (1/64) = 5/16. Now for the seventh ball to be white the probability is 1/2. So, total probability = (5/16)*(1/2) = 5/32 read less
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Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

the probability that 7th drawn ball is 4th white ball = 1/2
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Science teacher

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Headstart to assured 95+ score in math

It's a bit lengthy. I will post a pic of solution in my gallery.
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Spoken English, BTech Tuition, C Language, C++ Language, CAD etc., with 4 years of experience

P(A) = 12/24=1/2, P(B)= 12/24= 1/2, White ball is drawn 4th time in 7th draw i.e, already 6 draws are completed in which 3 white balls are already drawn. Required probability is: 6C3 p^3 q^3 p^1, where p is drawing white ball, q is drawing red ball. =5/32.
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Dear Akshay, The answer is too long and my answer reached the status of a research paper. But, however, I came to know this problem is of a category namely Markov's chain and so there was nothing new in my paper. Markov's chain come in because each draw (after the first one) depends on the outcomes...
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Dear Akshay, The answer is too long and my answer reached the status of a research paper. But, however, I came to know this problem is of a category namely Markov's chain and so there was nothing new in my paper. Markov's chain come in because each draw (after the first one) depends on the outcomes of the previous draws, as you said second ball drawn (and onwards) replaces the previously drawn ball (hence the previously drawn ball remains outside deciding the number of white and red balls from which the fresh drawing of a ball is done). Any way, thanks for the question as it improved my knowledge tremendously. read less
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Data Analytics & Advanced Statistics Tutor with 18 years of experience in teaching

Here P(white ball)=1/2 .Then required probability = 6C3 *(1/2)^4 *(1/2)^3 = 0.15625.(By Negative Binomial distribution)
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Probability is 1/2
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Solution: P(w) = 12/24=1/2, P(R)= 12/24= 1/2, White ball is drawn 4th time in 7th draw means, already 6 draws are completed in which 3 white balls are already drawn. Required probability is: 6C3 p^3 q^3 p^1, where p is drawing white ball, q is drawing red ball. =5/32
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