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For a random experiment, four red and four blue balls are available. In the beginning, four of the eight bullets are placed in one urn, the other four serve as a supply. In each step of the random experiment, a ball is now randomly drawn from the urn and exchanged for a ball of the other color from the supply

a) In the beginning, there are only red balls in the urn. With what probability is the fifth drawn ball red

b) In the beginning, there are now two red and two blue balls in the urn. With what probability are after 10 steps again two red and two blue balls in the urn

The answer for a) is 0 but for b) I'm not sure how to count it as $0.5^10 \times 0.25$ or? I just don't know how to do task b) if anyone could help me, thanks so much

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  • $\begingroup$ How could the answer for (a) be zero? $(4r,0b)\to$draw a red$\to (3r,1b)\to$ draw a red$\to(2r,2b)\to$draw a red$\to(1r,3b)\to$draw a blue$\to(2r,2b)\to$draw a red. How is that not a possible sequence of outcomes that occurs with nonzero probability? $\endgroup$
    – JMoravitz
    Commented Sep 21, 2019 at 17:01
  • $\begingroup$ Why do you say the answer for part a) is $0$? $\endgroup$
    – saulspatz
    Commented Sep 21, 2019 at 17:01
  • $\begingroup$ If u have 4 red balls in 1 urn then in other one you shoud have 4 blue, and how can u draw a red ball from the 4 blues? Or i missunderstood the questions a) $\endgroup$ Commented Sep 21, 2019 at 17:05
  • $\begingroup$ You seem to have completely misunderstood the question. We have an urn. In this urn we have four balls. We also have four balls in our pocket. There is only one urn. Now... we on each turn will grab a ball at random from the urn. We look at the color, we take note of what it was, we write it down in case we wanted to remember later, and then we reach into our pocket and grab a ball that was a different color than the one we just pulled out of the urn and then we place that ball into the urn and the one we just pulled goes into our pocket. $\endgroup$
    – JMoravitz
    Commented Sep 21, 2019 at 17:07
  • $\begingroup$ So, since we start with the urn containing only red balls, we will of course pull a red ball and we replace it with a blue ball from our pocket. Now... the second turn, we could have possibly grabbed a red ball or we could have possibly grabbed a blue ball. If we grabbed a red ball, we replace it with a blue ball from our pocket, now leaving the urn with two red and two blue balls, or it is possible that we grabbed a blue ball in which case we put the blue ball into our pocket and put a red ball back into the urn, leaving it with four red and zero blue. $\endgroup$
    – JMoravitz
    Commented Sep 21, 2019 at 17:08

1 Answer 1

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@JMoravitz is correct, Markov chains are a good way to go, but I assume you have never met them. It is not too hard to do this from first principles.

Here is part (a), starting with 4 R. Note that after an odd number of moves the number of R must be odd. In fact this must be the sequence:

RRRR $\to$ RRRB $\to$ RRBB or RRRR $\to$ RBBB or RRRB $\to$ BBBB or RRBB or RRRR $\to$ BBBR or RRRB.

Now assign probs. After 1 move, we have RRRB with prob 1. After 2 moves with have RRBB with prob 3/4 and RRRR with prob 1/4. After 3 moves we have RBBB with prob 3/8, RRRB with prob 5/8. After 4 moves with have BBBB with prob 3/32 (1/4 times 3/8), RRBB with prob 3/4 ($=3/4\times3/8+3/4\times5/8$), RRRR with prob 5/32 (=1/4 times 5/8).

So the probability that the fifth ball drawn is R is $1/2\times3/4+5/32=17/32$.

Part (b) involves 10 steps. It is perfectly feasible to adopt the approach above, but obviously fairly tiresome and error-prone. So maybe it is worth figuring out a more systematic approach. But you have given nil indication of how much you are expected to know.

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For (b), after 9 moves you must have 1 red or 3 reds. By symmetry the probabilities must be equal, so they must each be 1/2.

So the prob of 2 reds after 10 moves must be $1/2\times3/4+1/2\times3/4=3/4$. But you should have been able to figure that out for yourself!

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  • $\begingroup$ Thanks so much I think I understood it rn, will try to get part b alone, so basicially i have to repeat it with 10times? or is there any "faster" method? $\endgroup$ Commented Sep 21, 2019 at 17:56
  • $\begingroup$ shoudnt it be a) on the last move 1/2x3/4 + 3/32 (BBBB --> RBBB) (RRRR is 5/32 and cant get R) so a) shoud be 15/32? or maybe i got it wrong way $\endgroup$ Commented Sep 21, 2019 at 19:23
  • $\begingroup$ @16yokidtryingtolearnmath What do you mean by the "fifth drawn ball"? Do you mean the ball drawn from the urn, or the ball drawn from the supply? I assumed you meant the ball drawn from the urn. So you draw a R from the urn if (i) you had RRBB after the 4th move (prob $1/2\times3/4$) or (ii) you had RRRR after the 4th move (prob $1\times5/32$). $\endgroup$
    – almagest
    Commented Sep 21, 2019 at 20:02
  • $\begingroup$ Ye but for RRRR after 4th move i got 5/32 and for BBBB 3/32 and RRBB 3/4 so if its from Urn its 3/4 x 1/2 as u said and + 5/32 = 17/32 and if from supply its 3/4 x 1/2 + 3/32= 15/32, and thanks again u really helped me a lot $\endgroup$ Commented Sep 21, 2019 at 20:17

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