Question

A die has been "loaded" so that the probability of rolling any even number is

5

27

and the probability of rolling any odd number is

4

27

. (Assume the die is six-sided with each side numbered one through six.)

(a)

Find the following probabilities. Enter your probabilities as fractions.

Pr(2 ∩ even)	=
Pr(even)	=
Pr(2 | even)	=

What is the probability of rolling a 2, given that an even number is rolled?

(b)

Find the following probabilities. Enter your probabilities as fractions.

Pr(3 ∩ (3 or 6))	=
Pr(3 or 6)	=
Pr(3 | (3 or 6))	=

What is the probability of rolling a 3, given that a number divisible by 3 is rolled?

A bag contains 5 red balls and 7 white balls. Two balls are drawn without replacement. (Enter your probabilities as fractions.)

(a) What is the probability that the second ball is white, given that the first ball is red?


(b) What is the probability that the second ball is red, given that the first ball is white?


(c) Answer part (a) if the first ball is replaced before the second is drawn.

A red ball and 5 white balls are in a box. If two balls are drawn, without replacement, what is the probability of getting each of the following? (Enter your probabilities as fractions.)

(a) a red ball on the first draw and a white ball on the second


(b) 2 white balls


(c) 2 red balls

Answer 1

Dice problem

Even numbers are 2,4,6 and odd numbers are 1,3,5

P(even) = 3/6 = 0.5 or 50%

P(odd) =3/6 =0.5

a) P(2 and even) = 1/3 = 0.33

P( even) =3/6 =1/2 =0.5

P(getting a 2 given that it is rolled to be an even number)

= P( 2 and an even number) / P (even number)

= (1/6)/(3/6) = 1/3 or 0.33%

b) P( a number 3) = 1/6

P( a multiple of 3) = 2/6 = 1/3

P(a number 3 is rolled given that it is a multiple of 3)

= P ( getting a 3 and a multiple of 3)/P(multiple of 3)

= (1/6) / ( 2/6) = 1/2 = 0.5

Bag full of balls

a) P( second ball is white given that first ball was red)

= P( first ball is red and second ball is white) / P(first ball is red)

= ( 5/12)×(7/11) / (5/12)

= 7/11

b) P ( 2nd ball is red given that 1st was white ball)

= P(1st ball is white and 2nd ball is red) / P(1st ball is white)

= (7/12)×(5/11) / (7/12)

= 5/11

c) if 1st ball is replaced then

P ( 2nd ball is white given that 1st one was red)

= (5/12)×(7/12) / (5/12) = 7/12

Red ball and white balls problem

a) P ( 1st ball is red and second ball is white)

= (1/6)×(5/5) = 1/6

b) P( 2 white balls) = (5/6)×(4/5) = 2/3

c) P ( 2 red balls) = (1/6)×(0/5) = 0

Because there exists only one red ball.