Question

PLEASE ANSWER ALL THE FOLLOWING. WORK DOES NOT NEED TO BE SHOWN

1There are 9 yellow and 5 red cards. If you randomly pick 6 cards, what is the probability that 4 of them are yellow?

2For the weekend, you plan to go to a park, eat at a restaurant, and see a movie. There are 4 parks, 5 restaurants, and 6 movie choices. How many different possible combinations are there?

3Form a group of 6 people, three people are selected to form a committee. How many different ways are there to form the committee?

4In how many different ways that 6 seats in a row can be seated by six people?

5If you roll a fair die 4 four times. How many possible outcomes are there?

Answer 1

1)

Answer)

P(yellow) = (9)/(9+5) = 9/14

We need to use binomial formula here

ncr = n!/(r!*(n-r)!)

P(r) = ncr*(p^r)*(1-p)^n-r

Here n = 6 (total number of trials)

And r = 4 (we want 4 of them to be yellow)

P(4) = 6c4*(9/14)^4*(1-(9/14))^6-4

P(4) = 0.3267631631

2)

We have 4 parks, 5 restaurants and 6 movie choices

Every park can be with a restaurant

That is number of ways = 4*5

And every park and restaurant can be with any of the 6 movies

Number of ways = 6*4*5

Required answer is 120

3)

We need to use combination formula here

Ncr = n!/(r!*(n-r)!)

N = 6 and r = 3

6c3 = 6*5*4*3!/(3!*(6-3)!)

= 20

4)

First seat can be occupied by 6 people

Second seat can be occupied by 5 people

Third seat can be occupied by 4 people (as we are left with 4 people after first and second seat are occupied)

Number of ways = 6*5*4*3*2*1

= 720

5)

There are 6 sides in a die

When we roll it 4 times

Possible outcomes = 6*6*6*6

= 1296