Q.1 Two cards are drawn from a well shuffled deck of 52 cards
without replacement. Determine the probability of each event:
a) Both cards are spades.
P(first is a spade)*P(second is a spade)
(13/52)(12/51) = 1/17
b) The second card is a spade.
1/4 The probability of any randomly draw card is a spade is always 1/4.
The probability would change once we knew what the first card was, but
we have no way of knowing that.
c) At least one card is a red card.
1 - the probability that they are both from the 39 non-spades.
1 - (39/52)(38/51) = 15/34
d) The first card is a face and the second card is a diamond.
Case 1: The first is a diamond face card.
(3/52)(12/51) = 1/68
Case 2: The first is a non-diamond face card.
(9/52)(13/51) = 3/68
Answer = Their sum = 1/68 + 3/68 = 1/17
Q.2 One fair black die and one fair white die are rolled.
Use the sample space, shown below, to answer the following Questions:
We'll assume the number on the left of each ordered pair is the
black die and the one on the right of each ordered pair is the
white die. That won't make any difference in some of these
(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)
What is the Probability:-
a) The sum of the two dice is 9?
(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)
Those 4 ways colored red out of 36, or 4/36 = 1/9
b) That the sum of the two dice is a least 4?
All but the 3 rolls (1,1), (1,2), and (2,1).
That's 33 out of 36 or 33/36 or 11/12
c) That the Quotient of the two dice is 3?
Only (3,1) or (6,2). That's 2 ways out of 36 or 2/36 or 1/18
d) That the white die is three less than the black die?
That's (6,3), (5,2), (4,1). That's 3 out of 36 or 3/36 or 1/12.
e) Of getting an odd number on both dice?
Half the numbers on a die are odd. The probability that
the black die will be odd is 1/2, and the probability that
the white die is odd is also 1/2.
So probability = (1/2)(1/2) = 1/4
g) Of obtaining the same number on each die if it is known that
the sum of the two dice is at most 7?
Since we know that the sum is at most 7, we reduce the sample
space to:
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5)
(3,1) (3,2) (3,3) (3,4)
(4,1) (4,2) (4,3)
(5,1) (5,2)
(6,1)
Those 3 red ones out of 21. That's 3/21 = 1/7
Q.3 Five cards are dealt from a well-shuffled deck of 52 cards.
What is the Probability that the hand will contain:
a) 2 tens and 3 fives?
(4C2)(4C3) = (6)(4) = 24 ways out of 52C5
That's 24/2598960 = 1/108290
b) Exactly 3 twos and two other cards that are not twos?
(4C3)(48C2) = (4)(1128) = 4512 out of 52C5
That's 4512/2598960 = 94/54145
c) Three of a kind?(three cards the same and the other cards
not the same as the three of a kind and not the same as each other?
Choose the Rank of the three 13C1 = 13 ways
Choose the Suits of the three 4C3 = 4 ways
Choose the ranks of the other two 12C2 = 66 ways
Choose the suit of the lower ranking one of those two = 4C1 = 4 ways
Choose the suit of the higher ranking one of those two = 4C1 = 4 ways
That's (13)(4)(66)(4)(4)=54912 ways out of 52C5
That's 54912/2598960 = 88/4165
Edwin
