Statistics question...help

Question

The probability that Sandra gets a ride to work on any morning is 0.7. 
The probability that she gets a ride 3 times in a 5-day work week is , and the probability that she gets a ride at least 2 times in a 5-day workweek is .

Answer 1

3 times is interpreted as exactly three times

 

This would be a binomial problem

 

One instance of this would be RRRXX   (She got a ride the first 3 days but not the last 2)

 

It's probability is (.7)(.7)(.7)(.3)(.3) = .03087

 

However there is more than 1 way for her to get 3 rides out of 5

 

Here are the other ways:

 

RRXRX

RRXXR

RXRRX

RXXRR

....

There are a total of 10 ways but we do not want to have to list all the ways just to figure how many there are. This would be very exhausting for some problems.

 

So we can use the following formula instead N!/((N-X)!X!)    where N = total trials and X = the number of success

 

5!((2!)3!)  = 10

 

 

So in your problem we will multiply our orginal calculated probability of .03087 times 10

 

ANSWER = 10(.03087) = .3087 = 30.87%

 

To get this number 10, we can also look at Pascal's triangle on the 5th Row

 

and count over to the K+1 entry (3+1) = 4

 

1  5  10  10  5   1

 

The overall equation will look like:

 

 

N!/((N-X)!X!)(P)^(X)(1-P)^(N-X)

 

5!/(2!3!)(.7)³(.3)² =.3087

 

Now for the 2nd question

 

N remains the same but X has different values (2,3,4,5)  

 

Just do 4 calculations and add all your answers that you obtain

 

or do three calculations X = 0 , 1 or 2  add your answers and then subtract your sum from 1.

 

Direct message me if you have any questions.