Question

A box contains 4 red balls, 3 yellow balls, and 3 green balls. You will reach into the box and blindly select a ball, take it out, and then place it to one side. You will then repeat the experiment, without putting the first ball back. Calculate the probability that the two balls you selected include a yellow one and a green one.

3. Consider a binomially distributed random variable constructed from a series of 8 trials with a 60% chance of success on any given trial. Calculate the probability that there are more than 3 successes.

Answer 1

A box contains 4 red balls, 3 yellow balls, and 3 green balls.

The total number of balls in box is= 4+3+3=10 balls.

Therefore two balls are selected without replacement from 10 balls in box. The possible number of ways or Total sample space is:

n= ¹⁰C₂= 45

Let, event E as "the two balls you selected include a yellow one and a green one.

A box contains 3 yellow balls and 3 green balls.

Here we want selected two balls include a yellow one and a green one simultaneously, by using multiplication principal. The possible number of ways of selecting one yellow ball from 3 yellow balls and one green ball from 3 green balls is:

m = ³C₁*³C₁=9

P(E)=m/n=9/45=0.2

The probability that the two balls you selected include a yellow one and a green one is 0.2.

3) Here, the random variable x is binomial distributed.

X = is number of success

n= Number of tria =8

p=probability of success=60%=0.60

x~ B(n=8, p= 0.6)

The probability mass function of x is:

P (X=x)= ⁿ​​​​​​C_{​​​​​x} * p^{​​​​​​x} *(1-p)^n-x    x = 0,1,2,...n, p+q=1

P (X=x)= ⁸C_x * (0.6)^x *(0.4)^8-x    x = 0,1,2,...8

The probability that there are more than 3 successes.

P(X> 3)= P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)

=0.2322+0.2087+0.2090+0.08958+0.0168

P(X>3)=0.7263

The probability that there are more than 3 successes is 0.7263.