Question

a die is tossed four times. A success is getting 1 or 2 on a toss find the mean and variance of the number of success​

Answer 1

Let x be the random variable denoting the number of times an odd number (the number of successes) when a die is tossed twice.

Then x takes the values 0,1,2

Let P(X=0) be probability of getting no odd number (both times showing even).

∴P(X=0)=
6
3
​
×
6
3
​
=
4
1
​


Let P(X=1) be probability of getting odd number once.

∴P(X=1)=2C
1
​

6
3
​
×
6
3
​
=
6
6
​
×
6
3
​
=
2
1
​


Let P(X=2) be probability of getting odd number twice.

∴P(X=2)=
6
3
​
×
6
3
​
=
4
1
​


Thus the probability distribution of X is given by
X=x x=0 x=1 x=2
P(X=x)
4
1
​

2
1
​

4
1
​

We know that mean E(X)=∑x
i
​
P
i
​
=0×
4
1
​
+1×
2
1
​
+2×
4
1
​


∴E(X)=0+
2
1
​
+
2
1
​
=1

Thus mean E(X)=1.

We know that var(X)=E(X
2
)−[E(X)]
2


E(X
2
)=∑x
i
2
​
P
i
​
=0×
4
1
​
+1
2
×
2
1
​
+2
2
×
4
1
​


∴E(X
2
)=0+
2
1
​
+4×
4
1
​
=
2
3
​


Thus var(X)=
2
3
​
−[1]
2
=
2
3
​
−1=
2
1
​


Hence mean is 1 and variance is
2
1