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Sagot :
Answer:
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
[tex]12C_2=\frac{12\times11}{2}=66[/tex]We get the X=0,1,2
If there is no red ball,
Then X=0, we get
P(X=0) is,
[tex]P(X=0)=\frac{5C_0\times7C_2^{}}{12C_2}[/tex]we get,
[tex]P(X=0)=\frac{1\times7\times3}{6\times11}=\frac{21}{66}[/tex]If there is 1 red ball then X=1
we get,
P(X=1) is
[tex]P(X=1)=\frac{5C_1\times7C^{}_1}{12C_2}[/tex][tex]P(X=1)=\frac{5\times7}{66}=\frac{35}{66}[/tex]If there is 2 red balls then X=2
we get,
P(X=2) is
[tex]P(X=2)=\frac{5C_2\times7C^{}_0}{12C_2}=\frac{10}{66}[/tex]2)To find cumulative distribution function F(x)
[tex]F(x)=P(X\leq x)[/tex]when x=0
we get,
[tex]\begin{gathered} F(0)=P(X\leq0)=P(X=0) \\ F(0)=\frac{21}{66} \end{gathered}[/tex]When x=1 we get
[tex]F(1)=P(X\leq1)=P(X=0)+P(X=1)[/tex][tex]\begin{gathered} F(1)=\frac{21}{66}+\frac{35}{66}=\frac{56}{66} \\ F(1)=\frac{56}{66} \end{gathered}[/tex]When x=2 we get,
[tex]\begin{gathered} F(2)=P(X\leq2)=P(X=0)+P(X=1)+P(X=2) \\ F(2)=\frac{21}{66}+\frac{35}{66}+\frac{10}{66}=\frac{66}{66} \end{gathered}[/tex][tex]F(2)=1[/tex]
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