A box initially contains 1 RED ball and 2 WHITE balls all the same size. You select a ball at random and then return it to the box together with another ball of the same colour. The box now contains four balls. Again you select a ball at random from the box. Find the probability that both selected balls are the same colour.

What I think:

red/red + white/white = (1/3*2/4)+(2/3*3/4)

  • 1
    $\begingroup$ Are you doubting the answer, the method, or the presentation ? $\endgroup$ Oct 18, 2016 at 8:20
  • $\begingroup$ To quote Meat Loaf: "Two Out of Three Ain't Bad." But, do work on the presentation; it is important. $\endgroup$ Oct 18, 2016 at 11:48

2 Answers 2

  • The probability of taking 1 red ball in the 1st turn is $\frac{1}{3}$

Now we add another red ball in the sample,

  • The probability of taking 1 red ball in the 2nd turn is $\frac{2}{4}=\frac{1}{2}$

Hence,the P(taking red out red ball twice)=$$\frac{1}{3}\frac{1}{2}=\frac{1}{6}$$

lly,P(taking red out white ball twice)=$$\frac{2}{3}\frac{3}{4}=\frac{1}{2}$$

P(taking out red ball twice)+P(taking out white ball twice)=$$\frac{2}{3}$$

  • $\begingroup$ "taking red out white ball twice" ? $\endgroup$ Oct 19, 2016 at 0:04
  • $\begingroup$ typo fixed.Thanks for notifying. $\endgroup$
    – user237454
    Oct 19, 2016 at 10:28

That is almost okay.   Your presentation is lacking, but it is apparent you grasp the method and those are the right numbers to use.

This is the Law of Total Probability at work.

Letting $B_1,B_2$ stand for the colour of the first and second ball drawn; with $r,w$ the colour constants, then the sought probability is:

$$\begin{align}\mathsf P(B_1{=}B_2) ~=~& \mathsf P(B_1{=}r)\mathsf P(B_2{=}B_1\mid B_1{=}r)+\mathsf P(B_1{=}w)\mathsf P(B_2{=}B_1\mid B_1{=}w)\\[1ex]=~& \tfrac 13\tfrac 24+\tfrac 23\tfrac 34\\[1ex]=~& \tfrac 8{12} \\[1ex] =~& \tfrac 23\end{align}$$

Work on your presentation; it is important.


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