Question

You are bidding on five items available on an online shopping site. You think that for each bid you have a 35​% chance of winning​ it, and the outcomes of the five bids are independent events. Let X denote the number of winning bids out of the five items you bid on. Find the probabilities. a. Explain why the distribution of X can be modeled by the binomial distribution. b. Find the probability that you win exactly 3 bids. c. Find the probability that you win 3 bids or fewer. d. Find the probability that you win more than 3 bids.

Answer 1

a) For each bid, there are only two possible outcomes. Either you win, or you lose. Bids are independent of other bids. So we use the binomial distribution.

b) 18.11% probability that you win exactly 3 bids.

c) 94.59% probability that you win 3 bids or fewer.

d) 5.41% probability that you win more than 3 bids.

Explanation:

For each bid, there are only two possible outcomes. Either you win, or you lose. Bids are independent of other bids. So we use the binomial distribution. This is the answer for a.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

35​% chance of winning​ it

This means that
`p = 0.35`

Five bids

This means that
`n = 5`

b. Find the probability that you win exactly 3 bids.

This is P(X = 3).

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 3) = C_(5,3).(0.35)^(3).(0.65)^(2) = 0.1811`

18.11% probability that you win exactly 3 bids.

c. Find the probability that you win 3 bids or fewer.

`P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(5,0).(0.35)^(0).(0.65)^(5) = 0.1160`

`P(X = 1) = C_(5,1).(0.35)^(1).(0.65)^(4) = 0.3124`

`P(X = 2) = C_(5,2).(0.35)^(2).(0.65)^(3) = 0.3364`

`P(X = 3) = C_(5,3).(0.35)^(3).(0.65)^(2) = 0.1811`

`P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1160 + 0.3124 + 0.3364 + 0.1811 = 0.9459`

94.59% probability that you win 3 bids or fewer.

d. Find the probability that you win more than 3 bids.

Either you win 3 bids or fewer, or you win more than 3 bids. The sum of the probabilities of these events is 100%.

From c, 94.59% probability that you win 3 bids or fewer.

100 - 94.59 = 5.41

5.41% probability that you win more than 3 bids.