Question
Four balls are to be drawn without replacement from a box containing 8 red and 4 white balls. If X denotes the number of red balls drawn, then find the probability distribution of X.
✓
Answer
A, four balls are to be drawn without replacement and X denote the number of red balls drawn.
So, X is a random variable that can take values 0, 1, 2, 3 or 4.
Now,
P(X = 0) = P(All white balls) = P(WWWW) $=\frac{4}{12}\times\frac{3}{11}\times\frac{2}{10}\times\frac{1}{9}=\frac{1}{495},$
P(X = 1) = P(One of red balls and three white balls) = P(RWWW) + P(WRWW) + P(WWRW) + P(WWWR)
$=\frac{8}{12}\times\frac{4}{11}\times\frac{3}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{8}{11}\times\frac{3}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{3}{11}\times\frac{8}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{3}{11}\times\frac{2}{10}\times\frac{8}{9}$
$=4\times\frac{8}{495}=\frac{32}{495}.$
P(X = 2) = P(Two red balls and two white balls) = P(RRWW) + P(RWRW) + P(RWWR) + P(WWRR) + P(W
P(X = 3) = P(Three red balls and one white ball) = P(RRRW) + P(RRWR) + P(RWRR) + P(WRRR)
$=\frac{8}{12}\times\frac{7}{11}\times\frac{6}{10}\times\frac{4}{9}+\frac{8}{12}\times\frac{7}{11}\times\frac{4}{10}\times\frac{6}{9}+\frac{8}{12}\times\frac{4}{11}\times\frac{7}{10}\times\frac{6}{9}+\frac{4}{12}\times\frac{8}{11}\times\frac{7}{10}\times\frac{6}{9}$
$=4\times\frac{56}{495}=\frac{224 }{495}.$
P(X = 4) = P(All red balls) = P(RRRR) $=\frac{8}{12}\times\frac{7}{11}\times\frac{6}{10}\times\frac{5}{9}$
$=\frac{70}{495}$
So, the probability distribution of X is as follows:
So, X is a random variable that can take values 0, 1, 2, 3 or 4.
Now,
P(X = 0) = P(All white balls) = P(WWWW) $=\frac{4}{12}\times\frac{3}{11}\times\frac{2}{10}\times\frac{1}{9}=\frac{1}{495},$
P(X = 1) = P(One of red balls and three white balls) = P(RWWW) + P(WRWW) + P(WWRW) + P(WWWR)
$=\frac{8}{12}\times\frac{4}{11}\times\frac{3}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{8}{11}\times\frac{3}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{3}{11}\times\frac{8}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{3}{11}\times\frac{2}{10}\times\frac{8}{9}$
$=4\times\frac{8}{495}=\frac{32}{495}.$
P(X = 2) = P(Two red balls and two white balls) = P(RRWW) + P(RWRW) + P(RWWR) + P(WWRR) + P(W
P(X = 3) = P(Three red balls and one white ball) = P(RRRW) + P(RRWR) + P(RWRR) + P(WRRR)
$=\frac{8}{12}\times\frac{7}{11}\times\frac{6}{10}\times\frac{4}{9}+\frac{8}{12}\times\frac{7}{11}\times\frac{4}{10}\times\frac{6}{9}+\frac{8}{12}\times\frac{4}{11}\times\frac{7}{10}\times\frac{6}{9}+\frac{4}{12}\times\frac{8}{11}\times\frac{7}{10}\times\frac{6}{9}$
$=4\times\frac{56}{495}=\frac{224 }{495}.$
P(X = 4) = P(All red balls) = P(RRRR) $=\frac{8}{12}\times\frac{7}{11}\times\frac{6}{10}\times\frac{5}{9}$
$=\frac{70}{495}$
So, the probability distribution of X is as follows:
|
$\text{X}$
|
$0$
|
$1$
|
$2$
|
$3$
|
$4$
|
|
$\text{P}(\text{X})$
|
$\frac{1}{495}$
|
$\frac{32}{495}$
|
$\frac{168}{495}$
|
$\frac{224}{495}$
|
$\frac{70}{495}$
|
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