Mathematics, 19.06.2020 21:57 mrme41

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.

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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.

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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.

Answer from: Quest

127

whether it's diameter, or radius, i believe if i remember correctly from school, every amount has to always add up to 180. so if you take 53, and starting adding, long story short, taking 127 and adding it to 53, gives you 180.

Find the value of y. (note: the line passing through the center of the circle is a diameter.)

Answer from: Quest

seven students earned $833 by shoveling snow. if they share the money equally, how much money will each student get

answer: each student earns $119

step-by-step explanation:

to find the answer you have to divide 833/7 which is 119

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