Math marbles probability question?

2 Answers

• Deja Vu

  a) Altogether there are 5 + 9 + 6 = 20 marbles. Of these 5 are red.

  The probability of getting a red with the first draw is therefore 5/20

  Given that it is red, there now remain 19 marbles in the bag of which 4 are red. Therefore the probability of getting a red on the second draw = 4/19

  Given that it is a red, there now remain 18 marbles in the bag and 3 are red. The probability that the third is red = 3/18

  Given that the third is also red there now remain 17 marbles in the bag but only two red. The probability is therefore 2/17.

  Therefore the probability of drawing 4 red marbles =

  = 5/20 * 4/19 * 3/18 * 2/17

  = 1/969

  c) next as it is easier!

  Using the same methodology

  The probability that the first is NOT red = 15/20

  The probability that the 2nd is also not red = 14/19

  The probability that the 3rd is also not red = 13/18

  The probability that the 4th is also not red = 12/17

  The probability that all 4 are non-red (ie none are red) =

  15/20 * 14/19 * 13/18 * 12/17

  = 91/323

  b)

  Lets ASSUME that the first two are red and the last two are not.

  Following the same principle the probability RRNN = 5/20 * 4/19 * 15/18 * 14/17 = 35/969

  Note the first two are the odds of the red and the last two the non-reds.

  Now lets ASSUME that it goes Red, Non-Red, Red, Non red.

  The probability = 5/20 * 15/19 * 4/18 * 14/17

  As you can see this also comes to 35/969

  (The nominators are exactly the same numbers but in a different order, and the denominators are exactly the same.

  Therefore each named “arrangement” of 2 reds and 2 non reds = probability 35/969

  All we have to do now is calculate the number of arrangents of any 2 from 4. This is given by Combinations 4C2

  = 4! /2!2! = (4*3*2*1) / (2*2) = 6

  Those six arrangements are

  RRNN, RNRN, RNNR, NRRN, NRNR, NNRR

  Therefore the probability of getting EXACTLY two red = 6 * 35/969

  = 210 / 969

• danger isn’t my good experience the two, yet right here is going. 🙂 additionally, i think of you’re able to nicely be examining the question incorrectly. The final assertion says to make certain the risk once you pick ONE marble. a million. Yellow and striped. There are 4 “yellow and striped” marbles interior the bag, so which you have a 4/15 = .27 danger which you will %. one among those. 2. Yellow marble or marble w/stripes. There are 5 “yellow w/dots” marbles, 2 “purple w/stripes” marbles, and four “yellow w/stripes” marbles interior the bag, so which you have a (5 + 2 + 4)/15 = 11/15 = .seventy 3 danger of determining on a “yellow or striped” one. Now you’re choosing 2 marbles (one after the different). “without substitute” skill that once you %. the 1st marble, you do no longer positioned it lower back into the bag in the previous making your 2d determination. a million. 2 blue marbles. There are purely 2 blue marbles interior the bag. you have a 2/15 risk of determining on a blue one on your first determination. Now bear in mind, no substitute, so there at the instant are purely 14 marbles interior the bag and assuming the 1st one you picked is blue, purely between the the rest 14 marbles is blue. so which you have a a million/14 risk of choosing the blue one for the duration of your 2d determination. The danger of choosing 2 blue ones is (2/15)(a million/14) = .0095 danger. ok, and now you’re choosing 3 marbles (one after the different) WITH substitute… so once you pick a marble, you toss it lower back into the bag. a million. 3 marbles with dots. There are 7 marbles w/dots interior the bag (2 purple and 5 yellow). you have a 7/15 risk of choosing one among those for the duration of each and each of your attracts, so the risk of choosing 3 of them in a row is (7/15)(7/15)(7/15) = .10.