Question

1. Consider the experiment of first drawing one card from a single suit and then rolling a single die

a. How many simple events are possible ?

b. List the out sample space .

c. What is the probability of picking the same value card as die roll ?

d. What is the probability of picking a different number card than value rolled ?

e. What is the probability of obtaining a combined value of less than ten (assume that face cards are worth 10 and the ace is worth 11

Answer 1

solution:

a)The total No.of cards of a single suit n(s1) = 13 (Ace(A) , 2 , 3 ,4 ,5 ,6 ,7 ,8 ,9 ,King(K) ,queen (Q) ,Jack (J) )

The Total outcomes when a single die rolled n(s2) = { 1,2,3,4,5,6} = 6

Simple event: Simple event is a event where all the possibles outcomes are equally likely to occur

In the event of picking a card from single suit and then rolling a single die, all outcomes are equally likely to occur

so,No.of possible samle events n(S) = n(s1) * n(s2) = 13*6 =78

b) The sample space would be:

S = { (A,1),(A,2),(A,3),(A,4),(A,5),(A,6)

  (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

(7,1),(7,2),(7,3),(7,4),(7,5),(7,6)

( 8,1),(8,2),(8,3),(8,4),(8,5),(8,6)

(9,1),(9,2),(9,3),(9,4),(9,5),(9,6)

(10,1),(10,2),(10,3),(10,4),(10,5),(10,6)

(K,1),(K,2),(K,3),(K,4),(K,5),(K,6)

(Q,1),(Q,2),(Q,3),(Q,4),(Q,5),(Q,6)

(J,1),(J,2),(J,3),(J,4),(J,5),(J,6) }

C) Let D = event that picking same card as die roll

n(D) = 6

P( picking the same value card as die roll ) = P (D)

= n(D)/n(S)

= 6/78

= 0.077

d) Let D' = event that  a different number card than value rolled

probability of picking a different number card than value rolled = P(D')

= 1 - P(D)

= 1 - 0.077

= 0.923

e) Let A =11 , K=Q=J = 10

Let B = event that obtaining combined value less than 10

= {   (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5)

(5,1),(5,2),(5,3),(5,4)

(6,1),(6,2), (6,3),(7,1),(7,2),( 8,1) }

n(B) = 33

  probability of obtaining a combined value of less than ten = P(B)

= n(B) / n(S)

= 33 / 78

= 0.4231