Question

Two balls are drawn in succession without replacement from a box containing 5 red balls and 7 black balls. The possible outcomes and the values of the random variable X, where X is the number of red balls.i-Findtheprobabilitydistributionofthenumberofredballs.ii- Find cumulative distribution function (CDF).

Answer 1

Given that,

Two balls are drawn in succession without replacement from a box containing 5 red balls and 7 black balls.

The possible outcomes and the values of the random variable X, where X is the number of red balls.

1) To find the probability distribution of the number of red balls.

Total possible outcome is

`12C_2=(12*11)/(2)=66`

We get the X=0,1,2

If there is no red ball,

Then X=0, we get

P(X=0) is,

`P(X=0)=\frac{5C_0*7C_2^{}}{12C_2}`

we get,

`P(X=0)=(1*7*3)/(6*11)=(21)/(66)`

If there is 1 red ball then X=1

we get,

P(X=1) is

`P(X=1)=\frac{5C_1*7C^{}_1}{12C_2}`
`P(X=1)=(5*7)/(66)=(35)/(66)`

If there is 2 red balls then X=2

we get,

P(X=2) is

`P(X=2)=\frac{5C_2*7C^{}_0}{12C_2}=(10)/(66)`

2)To find cumulative distribution function F(x)

`F(x)=P(X\leq x)`

when x=0

we get,

`\begin{gathered} F(0)=P(X\leq0)=P(X=0) \\ F(0)=(21)/(66) \end{gathered}`

When x=1 we get

`F(1)=P(X\leq1)=P(X=0)+P(X=1)`
`\begin{gathered} F(1)=(21)/(66)+(35)/(66)=(56)/(66) \\ F(1)=(56)/(66) \end{gathered}`

When x=2 we get,

`\begin{gathered} F(2)=P(X\leq2)=P(X=0)+P(X=1)+P(X=2) \\ F(2)=(21)/(66)+(35)/(66)+(10)/(66)=(66)/(66) \end{gathered}`
`F(2)=1`