TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

P( getting 1 or 6)=p=‌

2

6

=‌

1

3

∴q=1−‌

1

3

=‌

2

3

Let X be the random variable showing the number of success as
∴P(X=0)=‌3C0(‌

1

3

)0(‌

2

3

)3=‌

8

27

P(X=0)=‌3C1(‌

1

3

)1(‌

2

3

)2=3×‌

1

3

×‌

4

9

=‌

4

9

P(X=2)=‌3C2(‌

1

3

)2(‌

2

3

)1=3×‌

1

9

×‌

2

3

=‌

2

9

P(X=3)=‌3C3(‌

1

3

)3(‌

2

3

)0=1×‌

1

27

×1=‌

1

27

∴ Hence, the probabilitydistribution of the number of six is given
∴ Mean =ΣXP(X)
‌‌=0×‌

8

27

+1×‌

4

9

+2×‌

2

9

+3×‌

1

27

‌‌=‌

4

9

+‌

4

9

+‌

1

9

=1
∴‌ Variance ‌=ΣX2P(x)−(‌ mean ‌)2
=02×‌

8

27

+(1)2×‌

4

9

+(2)2×‌

2

9

+(3)2×‌

1

27

−(1)2
=(‌

4

9

+‌

8

9

+‌

3

9

)−1=‌

15

9

−1=‌

6

9

=‌

2

3