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# Probability questions with cards, Tree diagrams vs Tables

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1. If an ace has a face value of one, a 2 has a value of two, a Jack has a value of 11, and so on for all of the cards, what is the probability of drawing a card that has a face value of less than six?

2.
A friend tells you that a tree diagram can be used to generate the sample space for any experiment, but a table or grid can not.
(a) Is this true?

(b) Use examples to explain your response in part (a).

3.
A card is drawn form a standard deck of playing cards, returned to the deck, and then a second card is drawn. Determine the following probabilities:
a. Both cards are aces or both cards are face cards.

b. Both cards are red or both cards are face cards

https://brainmass.com/math/probability/probability-questions-cards-tree-diagrams-versus-tables-295390

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1. If an ace has a face value of one, a 2 has a value of two, a Jack has a value of 11, and so on for all of the cards, what is the probability of drawing a card that has a face value of less than six?

There 13 possible values for cards.

Ace = 1.

2, 3, 4, 5, 6, 7, 8, 9, 10 for the numbered cards.

Jack = 11, Queen =12, King = 13.

Less than 6 means cards can be Ace or 2 or 3 or 4 or 5. Thus, there are 5 possible values.

Probability of face value less than 6 = 5/13

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2. A friend tells you that a tree diagram can be used to generate the sample space for any experiment, but a table or grid can not.
(a) Is this true?
No. This is not true. A tree diagram is only a convenient way to visualize all the outcomes in a sample space. This can also be done using a table but may be more tedious in some cases compared to the tree diagram.

(b) Use examples to explain your response in part (a).
When attempting to determine a sample space (the possible outcomes from an experiment), it is often helpful to draw a diagram which illustrates how to arrive at the answer. One such diagram is a tree diagram. In addition to helping determine the number of outcomes in a sample space, the tree diagram can be used to determine the probability of individual outcomes within the sample space. The probability of any outcome in the sample space is the product (multiply) of all possibilities along the path that represents that outcome on the tree diagram.

For example, consider the sample space for tossing a penny and a die.
(H for heads and T for tails in the penny each with probability = 1/2).
There are 6 possible numbers in a die namely, 1, 2, 3, 4, 5, 6 each with probability 1/6.

The tree diagram for this example is included in the file tree1.gif. The probability values are shown along the branches.

Note that it is always possible to enumerate all the paths in the tree diagram in a table.

In this case the Sample space is: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6 }

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3. A card is drawn form a standard deck of playing cards, returned to the deck, and then a second card is drawn. Determine the following probabilities:
a. Both cards are aces or both cards are face cards.

Note that there are 4 aces and a total of 52 cards in a pack. The chance that a random card drawn from the pack is an ace = 4/52 = 1/13

The chance that both cards are aces = 1/13 x 1/13 = 1/169

Now there are 12 face cards in a pack, namely 4 kings, 4 queens and 4 Jacks.

The probability of drawing a face card in a draw = 12/52 = 3/13

The probability of drawing face cards in both draws 3/13 x 3/13 = 9/169

Thus, the total probability that the cards are both aces or both face cards = 1/169 + 9/169 = 10/169

Note that drawing a face card and drawing an ace are mutually exclusive events. This means that they both
cannot happen together. Therefore, we simply add their probability values.

b. Both cards are red or both cards are face cards

The event in part b , can be split into 2 events that are mutually exclusive of one another (i.e. chance that they both happen = 0).

1. Both cards are red.
2. Both cards are face cards but not both red.

The probability of the event in b is simply the sum of the probabilities of these 2 mutually exclusive events.

1. Probability that both cards are red = 1/2 x 1/2 = 1/4

Note that there 26 red cards and 26 black cards in the pack. Thus, chance of a red card = 26/52 = 1/2.

2. Now there are 12 face cards in a pack and out of these 6 are black and 6 are red.

Thus:
Chance of drawing black face cards in both draws = 6/52 x 6/52 = 9/676

Chance of drawing black face card in first draw and red face card in second = 6/52 x 6/52 = 9/676

Chance of drawing red face card in first draw and black face card in second = 6/52 x 6/52 = 9/676

Total probability of drawing 2 face cards such that they are not both red = 27/676

Probability of event in b = 1/4 + 27/676 = 0.25 + 0.03994 = 0.28994 nearly.
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