1
$\begingroup$

i do not really understand the probability question and answer given.

Question: An urn contains 10 balls, of which three are red and seven are blue. Balls are randomly removed one at a time without replacement until all the red balls are drawn or all the blue balls are drawn. What is the probability that the last ball drawn is blue?

Solution

What i did was 1-(all 3 red balls drawn) = 1 -( 1/10 x 1/9 x 1/8 )

Which i understand that it is wrong. But i still can't seem to understand the solution. Could anyone kindly explain it to me? Thank you very much!

$\endgroup$

2 Answers 2

1
$\begingroup$

Stephen meskin gives a good approach to solve it

  1. All seven blue balls drawn before any red ball can be done in $1$ case only.

  2. 6 blue balls and 1 red ball be drawn before last blue ball. Whether ball chosen is red or blue will matter when withdrawing ball. That is BBBBBBR and BBBBBRB are not same. so there will be $\binom{7}{1}$ case.

  3. 6 blue balls and 2 red ball be drawn before last blue ball. Again order of chosen balls will matter. the number of case are $\binom{7}{2} + \binom{7}{1} = \binom{8}{2}$

number of favorable outcome is $1+7 +28 =36$

total number of outcome is $\binom{10}{3}$ (three position out of 10 for reds)

the probability is $\frac{36}{\binom{10}{3}} = 1/3$

$\endgroup$
1
$\begingroup$

There are three cases to consider.

  1. Seven blue balls are drawn before any red ones are drawn.
  2. Six blue balls and one red ball are drawn and then the eighth ball is blue.
  3. Six blue balls and two red balls are drawn and then the ninth ball is blue.

Compute the three probabilities and add them up.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .