Question

. A small town has 5600 residents. The residents in the town were asked whether or not they favored building a new bridge across the river. You are given the following information on the residents' responses, broken down by gender:

Men Women Total

In Favor 1400 280 1680

Opposed 840 3080 3920

Total 2240 3360 5600

A) What is the probability of a randomly selected resident being In a Woman?

B) What is the probability that a randomly selected resident is a Woman and is In Favor of the bridge?

C) What is the probability of a randomly selected resident being a Man or Opposed to the bridge?

D) If a randomly selected resident is a Woman, what is the probability that she is Opposed to the bridge?

E) Are gender and opinion about the bridge mutually exclusive events? Why?

F) Are gender and opinion about the bridge independent events? Why? Show some "proof" with probabilities.

2. How many Combinations of 4 students can be selected from a group of 10 students?

3. Describe the Sample Space for the experiment of selecting one card from a deck of regular playing cards?

Answer 1

a) probability of a randomly selected resident being In a Woman=3360/5600=0.60

b)

probability that a randomly selected resident is a Woman and is In Favor of the bridge=280/5600

=0.05

c)

probability of a randomly selected resident being a Man or Opposed to the bridge

=1-P(female and in favour)=1-0.05=0.95

d)probability that she is Opposed to the bridge given women=3080/3360=0.9167

e)

no as event being male and in favour contain some common elements therefore they can not be mutually exclusive; for mutually exclusive events N(gender and opinion) should be 0

f)

here as P(women )*P(in favour)=(3360/5600)*(1680/5600)=0.18 which is not equal to

probability that a randomly selected resident is a Woman and is In Favor of the bridge.

therefore  gender and opinion about the bridge are not independent events

2)

number of Combinations of 4 students out of 10 =¹⁰C₄ =10!/(4!*(10-4)!)=210

3)

here let card of spade, heart, club ; diamond are S,H,C,D

therefore sample space is {1S,2S,3S,...JS,QS,KS,2H,2S,..........QD,KD}

( whcih contain all 52 cards)