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Question: (2 points) Consider the experiment where a fair coin is flipped 3 times. Let F represent the event that Heads come up at most 2 in the three flips. Find P(F) 2. 3. Suzan grabs five marbles at random from a bag containing four red marbles, three green ones, two white ones and one purple one. What is the probabilities of the following events expressing each as a fraction in the lowest term a. She has at least 2 green ones. (2 points) b. She does not have all the red ones. (2 points)

Transcribed text From Image:(2 points) Consider the experiment where a fair coin is flipped 3 times. Let F represent the event that Heads come up at most 2 in the three flips. Find P(F) 2. 3. Suzan grabs five marbles at random from a bag containing four red marbles, three green ones, two white ones and one purple one. What is the probabilities of the following events expressing each as a fraction in the lowest term a. She has at least 2 green ones. (2 points) b. She does not have all the red ones. (2 points)

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2) Given data

A coin is tossed three times

so the sample space ={HHH ,HHT ,HTT, TTH, HTH, THT, THH ,TTT } =8

events in which head comes at most 2 times (F) ={HHT ,HTT ,TTH, HTH, THT ,THH, TTT}=7

7 P(F) 0.875

3)given data

Green marbel =4

red marbel =3

white marbel =2

purple marbel=1

total number of marbel =10

number of times marbel taken (n)=5

a) we have to find the probablity of getting at least 2 red marbel

let probablity of getting a red marble =R

here sucess getting a red marbel

so probablity of sucess (p)=4/10=0.4

probablity of failure (q)=1-p =1-0.4=0.6

now from the formula of binomial approximation

P(R\geq 2)=\frac{5!}{(5-2)!2!}(0.4)^{2}(0.6)^{(5-2)}+\frac{5!}{(5-3)!3!}(0.4)^{3}(0.6)^{(5-3)}+\frac{5!}{(5-4)!4!}(0.4)^{4}(0.6)^{(5-4)}+\frac{5!}{(5-5)!5!}(0.4)^{5}(0.6)^{(5-5)}=0.3456+0.2304+0.0768+0.0102=0.6630

b) now we have to find the probablity of getting 4 red ball at most

let probablity of getting a red marble =R

P(R\leq 4)=\frac{5!}{(5-0)!0!}(0.4)^{0}(0.6)^{(5-0)}+\frac{5!}{(5-1)!1!}(0.4)^{1}(0.6)^{(5-1)}+\frac{5!}{(5-2)!2!}(0.4)^{2}(0.6)^{(5-2)}+\frac{5!}{(5-3)!3!}(0.4)^{3}(0.6)^{(5-3)}+\frac{5!}{(5-4)!4!}(0.4)^{4}(0.6)^{(5-4)}=0.0778+0.2592+0.3456+0.2304+0.0768=0.9898


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