Let X be the number of accidents per week in a factory. Let the pmf of X be f (x) = 1 (x + 1)(x + 2) = 1 x + 1 − 1 x + 2 , x = 0, 1, 2, . . . . Find the conditional probability of X ≥ 4, given that X ≥ 1.

0.4

Step-by-step explanation:

The probability of X>=4 given X >=1 can be written:

Let A be X>=4.

Let B be X>=1.

The probability that A given B, P(A|B) equals...

P(A and B) / P(B).

So we need P(A and B) and P(B). P(A and B) is the same as P(A), since all X>=4 are also >=1.

Let's make a chart.

x f(x)

0 1([0]+1) - 1/([0] + 2) = .5

1 1([1]+1) - 1/([1] + 2) = .16667

2 1([2]+1) - 1/([2] + 2) = .08333

3 1([3]+1) - 1/([3] + 2) = .05

4 1([4]+1) - 1/([4] + 2) = .03333

5 1([5]+1) - 1/([5] + 2) = .02381

A is the P(X>=4), which is the complement of P(X<3).

P(X<3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) = .5 +.16667 + .08333 + .05 = .8. The complement = 1 - 0.8 = 0.2

B is the P(X>=1), which is the same as complement of P(X=0). 1-P(X=0) = 1-0.5 = 0.5

P(A) = 0.2 = 1/5

P(B) = .5 = 1/2

P(A|B) = 1/5 / 1/2 = 0.4

1d

2b

3 false

4 d

5 3200sq,ft

80+60+y=180

then divide the difference between 60 and y and add it to 180

so x =180 +( (y-60)/2)

x equals 200

step-by-step explanation:

8

since proposition states that a number is greater than 25, if is true then the number must actually be greater than 25.

this implies that all possible options, except 24, can be accepted