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The probability of blue ball is 0.5 and the probability of red ball is 0.3. I need the probability of pulling out 3 blue balls and one red ball (I always return the ball back inside). I have 4 options:

B B B R

B B R B

B R B B

R B B B

So my probability is 4 x (0.5)(0.5) (0.5)(0.3) I guess. My question is how can I find the number of options 4 in case it is too big and I can't find it just by listing my options. C might help cause 4c3 gives 4 and also 4c2 gives 6 which is correct but I don't know why.

Thank you!

Ori David
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  • The factor $\binom{n}{k}$ accounts for the number of ways exactly $k$ of the $n$ positions can be filled with a ball of a particular color. – N. F. Taussig Mar 07 '20 at 19:47

1 Answers1

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You can find the number of arrangements for your problem by using this method. To find the possible arrangements of n unique objects we use n!. However, if there is repetition like in this case, the number of arrangements are over estimated.

  • -like in this case
  • -4!=24
  • -so we need to cancel the effect of different arrangements of the similar type characters by dividing them
  • -like here we will divide 4! by 3! and 1!

  • -so the effect of 3Bs is canceled hence -4!/3!=4

  • The probabilty= 4 x (0.5)(0.5) (0.5)(0.3)

  • -if consider another example like

  • BBRR BRBR RRBB RBBR BRRB RBRB

  • -this is verified by the formula 4!/(2!*2!)=6 arrangements

  • as now there are two types

-Also, check out this link Similar Question

Ashar Jamal
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  • Thank you, I appreciate your help. I have known the formula $n! / k!(n-k)!$ I just had no clue what is does. Now I do. – Ori David Mar 07 '20 at 22:18
  • Good for you @Ori David. Please mark the question as answered if you feel satisfied by the answer. – Ashar Jamal Mar 08 '20 at 05:17