probability easy (400-520)

[Night reader]

If x is to be chosen at random from the set {1; 2; 3} and y is to be chosen at random from the set {4; 5; 6; 7}, what is the probability that xy will be even?

(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6

[Rahul@gurome]

If x is to be chosen at random from the set {1; 2; 3} and y is to be chosen at random from the set {4; 5; 6; 7}, what is the probability that xy will be even?

(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6

Total number of ways to select x and y = 3*4 = 12

Number of ways to select x and y such that xy is even,

• If x = 1 or 3, y can be 4 or 6 --> (2 + 2) = 4 ways
  If x = 2, y can be any of the four --> 4 ways

A total of 8 ways.

Required probability = 8/12 = 2/3

The correct answer is D.

[Night reader]

If x is to be chosen at random from the set {1; 2; 3} and y is to be chosen at random from the set {4; 5; 6; 7}, what is the probability that xy will be even?

(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6

Total number of ways to select x and y = 3*4 = 12

Number of ways to select x and y such that xy is even,

• If x = 1 or 3, y can be 4 or 6 --> (2 + 2) = 4 ways
  If x = 2, y can be any of the four --> 4 ways

A total of 8 ways.

Required probability = 8/12 = 2/3

The correct answer is D.

Thanks Rahul, that must be too easy for you.

[apex231]

I am wondering shouldn't it be 5/6?

P to select 2 = 1/3
P of even product = 1/3 * 1 = 1/3

P to select 4 = 1/4
P of even product = 1/4 * 1 = 1/4

P to select 6 = 1/4
P of even product = 1/4 * 1 = 1/4

Total P of even product = 1/3 + 1/4 + 1/4 = 5/6


Also, using combinations
when 2 is selected, no. of even products = 1 * 4 = 4
when 4 is is selected, no. of even products = 1 * 3 = 3
when 6 is is selected, no. of even products = 1 * 3 = 3

Total even products = 4+3+3 = 10
Total products = 3*4 = 12

Probability of even product = 10/12 = 5/6

Am i missing something?

[apex231]

I am wondering shouldn't it be 5/6?

P to select 2 = 1/3
P of even product = 1/3 * 1 = 1/3

P to select 4 = 1/4
P of even product = 1/4 * 1 = 1/4

P to select 6 = 1/4
P of even product = 1/4 * 1 = 1/4

Total P of even product = 1/3 + 1/4 + 1/4 = 5/6


Also, using combinations
when 2 is selected, no. of even products = 1 * 4 = 4
when 4 is is selected, no. of even products = 1 * 3 = 3
when 6 is is selected, no. of even products = 1 * 3 = 3

Total even products = 4+3+3 = 10
Total products = 3*4 = 12

Probability of even product = 10/12 = 5/6

Am i missing something?

Got it. I am counting combinations with 2 twice....