.A card is drawn from a standard deck of 52 playing cards. Find the probability of
each of the following events occurring:
a) The card is a club.
b) The card is a five.
c) The card is a red queen.
d) The card is not a face card.
For a) I got 13/52...since there are 13 cards in a suit. Am I doing this correctly?
Given that two six-sided dice are rolled once:
a) What are the odds in favour of rolling a sum of 9?
b) What are the odds against rolling a sum that is a multiple of 3?
each of the following events occurring:
a) The card is a club.
b) The card is a five.
c) The card is a red queen.
d) The card is not a face card.
For a) I got 13/52...since there are 13 cards in a suit. Am I doing this correctly?
Given that two six-sided dice are rolled once:
a) What are the odds in favour of rolling a sum of 9?
b) What are the odds against rolling a sum that is a multiple of 3?
-
Yes that is the correct way to go about answering them:
When the answers are short there is little need for lots of notation but a general notation when answering probability questions is to write P(x = y).
Meaning the probability that x (the action taking place, in this case x is the card you pull out) happening to be y (a certain possibility, in this case the card you are looking for)
1.a) P(x = club) = 13/52
b) P(x = 5) = 4/52 = 1/13
c) P(x = Red Queen) = 2/52 = 1/26
d) P(x = Face Card) = 36/52 = 9/13
2. a) P(x = sum of 9) = 4/36 = 1/9
This is because there are 36 possible combinations the dice can make but only 4 combinations (6,3) (3,6) (5,4) (4,5) that add up to 9.
b) P(x = sum is not multiple of 3) = 24/36 = 2/3
This time the possible combinations are (1,2) (2,1) (5,1) (1,5) (4,2) (2,4) (3,3) (6, 3) (3,6) (5,4) (4,5) (6,6)
This would give 12/36 but you want the odds that it isn't a multiple of 3 so you need to do [1 - 12/36] to find the answer
When the answers are short there is little need for lots of notation but a general notation when answering probability questions is to write P(x = y).
Meaning the probability that x (the action taking place, in this case x is the card you pull out) happening to be y (a certain possibility, in this case the card you are looking for)
1.a) P(x = club) = 13/52
b) P(x = 5) = 4/52 = 1/13
c) P(x = Red Queen) = 2/52 = 1/26
d) P(x = Face Card) = 36/52 = 9/13
2. a) P(x = sum of 9) = 4/36 = 1/9
This is because there are 36 possible combinations the dice can make but only 4 combinations (6,3) (3,6) (5,4) (4,5) that add up to 9.
b) P(x = sum is not multiple of 3) = 24/36 = 2/3
This time the possible combinations are (1,2) (2,1) (5,1) (1,5) (4,2) (2,4) (3,3) (6, 3) (3,6) (5,4) (4,5) (6,6)
This would give 12/36 but you want the odds that it isn't a multiple of 3 so you need to do [1 - 12/36] to find the answer
-
a) The card is a club. ok !!
b) The card is a five. 4/52
c) The card is a red queen. 2/52
d) The card is not a face card. 40/52
Given that two six-sided dice are rolled once:
a) What are the odds in favour of rolling a sum of 9?
favorable: 3-6,4-5,5-4,6-3, ie 4 ways, unfavorable = 36-4 = 32
odds in favor = 4:32 = 1:8
b) What are the odds against rolling a sum that is a multiple of 3?
favorable: 1-2,2-1//1-5,2-4,3-3,4-2,5-1//3-6,4-5,5-… = 12
unfavorable = 36-12 = 24
odds against = 24:36 = 2:3
b) The card is a five. 4/52
c) The card is a red queen. 2/52
d) The card is not a face card. 40/52
Given that two six-sided dice are rolled once:
a) What are the odds in favour of rolling a sum of 9?
favorable: 3-6,4-5,5-4,6-3, ie 4 ways, unfavorable = 36-4 = 32
odds in favor = 4:32 = 1:8
b) What are the odds against rolling a sum that is a multiple of 3?
favorable: 1-2,2-1//1-5,2-4,3-3,4-2,5-1//3-6,4-5,5-… = 12
unfavorable = 36-12 = 24
odds against = 24:36 = 2:3
-
A---13 clubs in deck so 13/52 or 1/4
B---4 "5's" in deck so 4/52 or 1/13
C---2 red queens so 2/52 or 1/26
D----12 face cards in deck of 52 so 40/52 or 10/13 not a face card
2 dice being rolled there are 36 different combinations 6 x 6 =36
write down all the combinations like 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 4.45 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, look for all the ones that add up to 9 with two dice there are 4 out of 36 combinations or 1/9
for the multiple of 3 you have to look at all the combinations that equal 3,6,9 or 12 in this case 1.2,2.1,1.5,2.4.3.3,4.2,5.1,3.6,4.5,5.4,… or 12 different combinations so 12/36 or 1/3
B---4 "5's" in deck so 4/52 or 1/13
C---2 red queens so 2/52 or 1/26
D----12 face cards in deck of 52 so 40/52 or 10/13 not a face card
2 dice being rolled there are 36 different combinations 6 x 6 =36
write down all the combinations like 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 4.45 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, look for all the ones that add up to 9 with two dice there are 4 out of 36 combinations or 1/9
for the multiple of 3 you have to look at all the combinations that equal 3,6,9 or 12 in this case 1.2,2.1,1.5,2.4.3.3,4.2,5.1,3.6,4.5,5.4,… or 12 different combinations so 12/36 or 1/3