Chapter Probability Part:2|Class 12|UGC CBCS| Statistics

                    Click here: Chapter Probability Part:1|Class 12|UGC CBCS| Statistics

Q11.Show that if a fair die is thrown twice getting 'five' in the first throw and getting 'six in the second throw are statistically independent.

Ans:

$\small A:$ getting five in the first throw. $\small P(A)=\frac{1}{6}$

$\small B:$ getting six in the second throw.$\small P(B)=\frac{1}{6}$

When a fair die throw twice the all possible out comes are 36.

(1,1),(1,2),......,(1,6)

(2,1),(2,2),.......,(2,6)

.

.

(5,1),(5,2),..(5,5),(5,6)

(6,1),(6,2),........,(6,6)

$\small P(A\cap B)=\frac{1}{36}=\frac{1}{6}.\frac{1}{6}=P(A)P(B)$

Q12.Each coefficient in the quadratic equation $\small ax^2+bx+c=0$ is determined by selecting randomly integers from 1,2,3,4,5. Find the probability that the equation will have 

i.equal roots 

ii. real roots

Ans: $\small ax^2+bx+c=0$  will have real roots if $\small b^2=4ac$

                                                                 equal roots if$\small b^2\geq 4ac$

$\small a,b,c$ can take 1,2,3,4,5 these five values 

So all possible outcomes are $\small 5^3=125$

Note: As we know the highest value of b=5, thats why $\small b^2_{max}=25$

a	c	4ac	values of b statisfies $\small b^2\geq 4ac$	cases statisfies $\small b^2\geq 4ac$	values of b statisfies $\small b^2= 4ac$	cases statisfies $\small b^2= 4ac$
1	1 2 3 4 5	4 8 12 16 20	2,3,4,5 3,4,5 4,5 4,5 5	4 3 2 2 1	2 4	1 1
2	1 2 3 4 5	8 16 24 X X	3,4,5 4,5 5	3       2       1	4	1
3	1 2 3 4 5	12 24 X X X	4,5 5	2 1
4	1 2 3 4 5	16 X X X X	4,5	2	4	1
5	1 2 3’ 4 5	20 X X X X	5	1

Total:                                               24                              4

$\small P(real-roots)$=$\frac{24}{125}$

$\small P(equal-roots)$=$\frac{4}{125}$

Q13.Suppose the chances of simultaneous occurence of two independent events A and B is $\small \frac{1}{6}$ and the probability that none of these two events occurs in $\small \frac{1}{3}$.Find $\small P(B)=?$

Let, P(A)=x

P(B)=y

$\small P(A\cap B)=xy=\frac{1}{6}$

=$\small xy=\frac{1}{6}$

$\small P(A^c \cap B^c)=\frac{1}{3}$

=$\small (1-x)(1-y)=\frac{1}{3}$

Now put $\small x=\frac{1}{6y}$

$\small y=\frac{1}{3},\frac{1}{2}$

Click here: Chapter Probability Part:3|Class 12|UGC CBCS| Statistics 

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