Question

In a game, players spin a wheel with ten equally likely outcomes, 1-10. Find the probability that:

(a) In a single spin, the number that comes up is odd AND less than 5.

(b) In a single spin, the number that comes up is odd OR less than 5.

(c) In two spins, an odd number comes up first and a number less than 5 comes up second.

Answer 1

Number of outcomes when the wheel with ten equally likley outcomes  is spun = 10 (1 to 10)

(a)

A: Event of Number that comes up is odd in a single spin

Outcomes that favor A : 1,3,5,7,9

Number of outcomes that favor A = 5

P(A) = probability that in a single spin, the number that comes up is odd =5/10=1/2

B : Event of Number that comes up is less than 5 a single spin

Outcomes that favor B : 1,2,3,4

Number of outcomes that favor B = 4

P(B) = probability that in a single spin, the number that comes up is less than 5 = 4/10=2/5

A and B : Event of Number that comes up is odd and less than 5 In a single spin

Events that favor A AND B = (1,3)

Number of outcomes that favor A AND B = 2

probability that in a single spin, the number that comes up is odd AND less than 5 = P(A AND B) = 2/10 =1/5

probability that in a single spin, the number that comes up is odd AND less than 5 = 1/5 = 0.2

(b)

Probability that In a single spin, the number that comes up is odd OR less than 5 = P(A OR B)

By addition theorem,

P(A OR B) = P(A)+P(B)-P(A AND B) = 1/2 + 2/5 - 1/5 = (5+4-2)/10 = 7/10

Probability that In a single spin, the number that comes up is odd OR less than 5 = 7/10 = 0.7

(c)

probability that in two spins, an odd number comes up first and a number less than 5 comes up second.

A1: Event of odd number comes up in the first spin

Outcomes that favor A1 : 1,3,5,7,9

Number of outcomes that favor A1 = 5

P(A1) = probability that odd number comes in the first spin = 5/10 =1/2

B2: Event of Number that comes up is less than 5 in the second spin;

Outcomes that favor B2 : 1,2,3,4

Number of outcomes that favor B2 = 4

P(B2) = 4/10 = 2/5

The outcome of the second spin does not depend on the outcome of first spin hence A1 and B2 are indpendent

Therefore, P(A1 AND B2) = P(A1)P(B2)=(1/2)(2/5) = 2/10 =1/5

probability that in two spins, an odd number comes up first and a number less than 5 comes up second= 1/5=0.2