Question

Two insurers provide bids on an insurance policy to a large company. The bids are between $1,900 and $2,300. The company decides to accept the lower bid if the two bids differ by $20 or more. Otherwise, the company will consider the two bids further. Assume that the two bids are independent and are both uniformly distributed on the interval from $1,900 to $2,300.

Assume that one of the insurers provided a bid of $2,100 (and you don’t know the bid provided by the other). Compute the probability that the company will accept one of the bids.

Please solve it algebraically, without referencing graphing.

Answer 1

Solution:

The company decides to accept the lower bid if the two bids differ by $20 or more. One insurers provided a bid of $2,100. Now compny will accept the bid of $2100 of first investor, if second investor has provided the bid of $2100+$20 = $2120 or more. Or company will accept the bid of second investor if second investor has provided the bid of $2100 - $ 20 = $2080 or less.

Let X represents the amount of bid provided by the second customer.

Hence, we have to obtain the probability that second investor has provided the bid of ($2120 or more) or ($2080 or less).

i.e. We need to obtain P(X ≤ $2080 or X ≥ 2120).

P(X ≤ $2080 or X ≥ 2120) = P(X ≤ $2080) + P(X ≥ 2120).

Given that , X is uniformly distributed in the interval from $1900 to $2300.

Hence, probability density function of X will be as follows:

{f(x) = 2300_1900 = do if 1900 SX S 2300

We shall obtain P(X ≤ $2080).

2080 P(X 5 2080) = f(2).de J1900

2080 1 P(X < 2080) = 2080 1.dc J1900 J1900 400

P(X 5 2080) = 100 [2]2900 = 10 [2080 – 1900) 400

P(X < 2080) = 0.45

Now we shall obtain P(X ≥ 2120).

2300 P(X > 2120) = | f(x).dx J2120

1 P(X > 2120) = 12300 1 J2120 400 2300 1.de 400 2120

P(X > 2120) = 100 [2]2120 = 100 [2300 – 2120) 400

P(X > 2120) = 0.45

Hence, P(X ≤ $2080 or X ≥ 2120) = 0.45+0.45 = 0.90

Hence, the probability that the company will accept one of the bids is 0.90.