Question

The random variable X, representing the number of defective laptops purchased by an office from a shipment of 20 computers, has the probability distribution function , x ? C C C f ( x ) x r x r     5 20 5 20 a) Assign values to x b) Find its cumulative distribution function and probability of purchasing 3 defective laptop. 

Answer 1

Given :-

random variable x = number of defective piece of laptop

total piece of laptop n =20

defective piece of laptop r =5

purchase laptop = 3

Experiment: selecting 3 computers at random out of 20 N(S) = ²⁰c₃ = 1140equally likely outcomes

The possible values of X are: x = 0, 1, 2, 3.

N(X=0)={0D and 3N} = ⁵c₀ .¹⁵c₃ = 1 × 455 = 455

N(X=1)={1D and 2N} = ⁵c₁ .¹⁵c₂ = 5 × 105 =525

N(X=2)={2D and 1N} = ⁵c₂ .¹⁵c₁ = 10 × 15 =150

N(X=3)={3D and 0N} = ⁵c₃ .¹⁵c₀ = 10 × 1 =10

P(X=0)=$\frac{455}{1140}$

P(X=1)=$\frac{525}{1140}$

P(X=2)=$\frac{150}{1140}$

P(X=3)=$\frac{10}{1140}$

The probability distribution of X  

X                     0            1           2           3             total
p(x) = P(X=x)        455/1140     525/1140     150/1140     10/1140        1

probability distribution function is:-

$$P(x) = \frac{{\dbinom{5}{x}}{\dbinom{15}{3-x}}}{\dbinom{20}{3}}$$

assign values of x

X                     0            1           2           3             total
p(x) = P(X=x)        455/1140     525/1140     150/1140     10/1140        1

cumulative distribution function

The cumulative distribution function (cdf) F(x) of a discrete rv variable X with pmf p(x) is defined by For any number x, F(x) is the probability that the observed value of X will be at most x.

$$F(x)=P(X≤x)=\sum_{y:y≤x}P(y)$$

probability of purchasing 3 defective laptop

$$F(3)=P(X=3)\\=\frac{10}{1140}$$