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Conditonal probability

suvadip

Member
Feb 21, 2013
69
A box contains 5 red and 10 white balls. Two balls are drawn at random without replacement. What is the probability that first ball drawn is red given that the second one is white? I am confused how the colour of the second ball effects the probability of the first ball. Please help.
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
Re: conditonal probability

A box contains 5 red and 10 white balls. Two balls are drawn at random without replacement. What is the probability that first ball drawn is red given that the second one is white? I am confused how the colour of the second ball effects the probability of the first ball. Please help.
You want to find \(\displaystyle \mathcal{P}(R_1|W_2)=\frac{\mathcal{P}(R_1W_2)}{ \mathcal{P}(W_2)}\)

But you know \(\displaystyle \mathcal{P}(W_2)=\mathcal{P}(R_1W_2)+\mathcal{P}(W_1W_2)\)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Rather than using memorized formulas, you can think it out this way:

There are 5 red balls and 10 white balls, a total of 15 balls. The probability that a red ball is drawn first is 5/15= 1/3. If that happens, there are still 10 white balls but now only 14 balls total. The probability the next ball drawn is white is 10/14= 5/7. The probability "the first ball is red and the second is white" is (1/3)(5/7)= 5/21.

The probability that a white ball is drawn first is 10/15= 2/3. If that happens there are now 9 white balls and 14 balls total. The probabilty that second ball drawn is white is 9/14. The probability "the first ball drawn is white and the second is white" is (2/3)(9/14)= 3/7.

The total probability for "white ball drawn second" is 5/21+ 3/7= 5/21+ 9/7= 14/21= 2/3. The probability "first ball is red given that the second ball is white" is (5/21)/(2/3)= (5/21)(3/2)= 5/14. The probability "first ball is white given that the second ball is white" is (3/7)/(2/3)= (3/7)(3/2)= 9/14.