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In a bag we have 5 red marbles and 3 white

We assume they are identical We pick from the bag randomly 3 marbles at the same time

  1. Calculate the probability of the events:

    -A:"picking three reds"

    -B:"picking at least one white "

  2. We replace the white marbles in the bag with $n$ Black marbles ($n$ > 1) We pick 2 marbles (one after one) without replacement

    a) calculate the probability of picking red at the second picking

    b) if the second marble picked is red, what is the the probability that the first one is black

This is my trying

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    $\begingroup$ What have you tried? This is pretty standard conditional probability. $\endgroup$ – Don Thousand Mar 25 '19 at 23:02
  • $\begingroup$ @DonThousand the first question is easy , 2.b) I have problem in calculating intersections of events $\endgroup$ – Abdennour Harche Mar 25 '19 at 23:11
  • $\begingroup$ If you post what you have tried, we may see what the trouble might be. But here's a hint: If the second marble is red, then the first marble could not have been that particular red marble. $\endgroup$ – Graham Kemp Mar 26 '19 at 0:24
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First, picking 3 marbles "at the same time" is the same as picking three marbles, one at a time, without replacement. Since there are 5 red and 3 white marbles, the probability the first marble chosen is red is 5/8. There are then 4 red and 3 white marbles left. The probability the second marble chosen is red is 4/7. There are then 3 red and 3 white marbles left. The probability the third marble chosen is red is 3/6= 1/2. The probability all three marbles are red is (5/8)(4/7)(1/2)= 5/28.

The probability of "at least one white" is 1 minus the probability of "all red" so 1- 5/28= 23/28.

For problem 2, if I am reading it correctly, we have 5 red marbles and n black marbles for a total of n+ 5 marbles. Now we need to consider two possible scenarios: 1) The first marble chosen is red. The probability of that is 5/(n+5). In this case, there are now 4 red and n black marbles for a total of n+ 4. In this scenario, the probability the second marble chosen is red is 4/(n+4).

2) The first marble chosen is black. The probability of that is n/(n+5). In this case, there are now 5 red and n- 1 black marbles so still a total of n+ 4. In this scenario, the probability the second marble is chosen is red is 5/(n+ 4).

Putting those together the probability that the second marble chosen is red is [5/(n+5)][4/(n+4)]+ [n/(n+ 5)][5/(n+ 4)]= 20/(n+5)(n+4)+ 5n/(n+5)(n+ 4)= (5n+ 20)/(n+5)(n+4)= 5(n+ 4)/(n+5)(n+4)= 5/(n+ 4).

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