A bag contains 8 red balls and some white balls. If the probability of drawing a white ball is half of the probability of drawing a red ball then find the probability of drawing a red ball and a white ball if the balls are drawn without replacement.
a
\(\frac {1}{3}\)
b
\(\frac {2}{9}\)
c
\(\frac {2}{3}\)
d
\(\frac {8}{33}\)
Explanation
Correct Option
dVideo Explanation
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Chrisplendour
2 weeks ago
calm down you'll understand it.
Red is 8.
let white be =n
P(Total)=8+n
P(Red)=8/(8+n)
P(white)=n/(8+n)
let's try getting number of white,n, using the information given.
FROM QUESTION=>P(W)=1/2P(R)
i.e white is half of red
n/8+n=1/2*(8/8+n)
n/8+n=4/8+n
n(8+n)=4(8+n)
(8+n )cancel out
n=4
white=4
red=8
total=12
red then white are picked without replacement.
red=8/12
white=4/11
8/12*4/11=8/33.
