Probability & Combinatorics Flashcards

1
Q

There are:

3 green marbles

2 red marbles

1 yellow marble

What is the probability of choosing a green OR yellow marble?

What does OR translate to?

A

There are 6 total marbles, so denominator is 6

OR translates to +

So, we add: 3/6 chance of green + 1/6 chance of yellow = 4/6 = 2/3

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2
Q

There are:

3 green marbles

2 red marbles

1 yellow marble

If we choose 2 marbles without replacement,

what is the probability of choosing a green AND a red?

What does AND translate to?

A

AND means we multiply the individual probabilities

There are 2 ways to get a green and a red:

A) 1st marble green, 2nd marble red:

3/6 * 2/5 = 1/2 * 2/5 = 1/5 (once we’ve picked one, there’s only 5 left)

B) 1st marble red, 2nd marble green:

2/6 * 3/5 = 1/3*3/5 = 1/5

Overall, we have (Green AND Red) OR (Red AND Green), so we add:

1/5 + 1/5 = 2/5

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3
Q

There are:

3 green marbles

2 red marbles

1 yellow marble

If we choose 2 marbles without replacement,

what is the probability of choosing 2 greens?

A

Choosing 2 greens means Green AND Green

AND means we multiply the individual probabilities

3/6 * 2/5 = 1/2 * 2/5 = 1/5

(after we’ve chosen one green, there’s only 2 out of 5 left)

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4
Q

There are:

3 green marbles

2 red marbles

1 yellow marble

If we choose 2 marbles with replacement,

What is the probability of choosing a yellow first, and then a red?

A

This one is with replacement, so the denominator stays at 6.

There is only 1 option, since it says yellow is first. (we can’t have red, then yellow)

1/6 * 2/6 = 1/18

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5
Q

There are 3 dogs and 2 cats.

If two are chosen at random, what is the probability that they will both be the same type?

A

Translate: AND is *, OR is +

2 ways: both dogs, OR both cats

(Dog AND Dog) OR (Cat AND Cat)

3/5 * 2/4 + 2/5 * 1/4 = 3/10 + 1/10 = 4/10

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6
Q

If we flip a coin 4 times, what is the probability of it landing on the same side on all 4 flips?

A

We could have all heads, OR all tails, so we add these 2 probabilities:

A) All heads: 1/2 * 1/2 * 1/2 * 1/2 = 1/16

B) All tails: 1/2 * 1/2 * 1/2 * 1/2 = 1/16

1/16 + 1/16 = 1/8

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7
Q

If there is a 40% chance of rain,

What is the probability that it will NOT rain on the first day, and rain on the second day

Express as a fraction

A

Probability of NOT A = 1 - Probability of A

Probability of NOT rain = 1 - 40% = 1 - 2/5 = 3/5

AND translates to multiply:

Not Rain * Rain = 3/5 * 2/5 = 6/25

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8
Q

There are 2 sets of cards with numbers:

Set A {1, 2, 4, 6, 9} and Set B {1, 2, 3, 5, 6, 8}

If Joe randomly chooses 1 card from each set,

what is the probability that the numbers multiplied together will be an even number?

A

There are 3 ways to get an even:

Even * Even

Odd * Even

Even * Odd

There’s only 1 way to get odd: Odd * Odd = Odd

So, it is easier to do 1 - Probability of Odd

Set A: 2/5 chance to get Odd (1 or 9)

Set B: 3/6 = 1/2 chance to get Odd (1, 3, or 5)

Chance of Odd AND Odd = 2/5 * 1/2 = 1/5

1 - 1/5 = 4/5

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9
Q

Two dice are rolled. If the sum of the numbers on the dice was 7, what is the probability that one of the dice showed a number greater than 4?

A

The possibilities are limited to those that give a sum of 7. (“IF the sum…)

Without this condition, there would be 36 total possibilities (6 for first die * 6 for second die)

List the possibilities:

1-6, 2-5, 3-4, 4-3, 5-2, 6-1

There are 6 total –> that’s the denominator.

4 have a number greater than 4.

4/6 = 2/3

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10
Q

How many different license plates can be made,

using 3 DISTINCT letters,

choosing from the letters A, B, C, D, E, F, G ?

A

Here, the order matters, so we don’t divide by 3! (3 factorial)

Write out 3 “Slots”, one for each letter.

___ ___ ___

There are 7 possibilites for the first slot, then 6 possibilites for the 2nd slot, then 5 for the 3rd slot (because the letters must be distinct).

We multiply them to find the total number of ways.

_7_ * _6_ *_5_ = 210

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11
Q

How many different teams of 3 can be made from 7 people?

A

35

When we are picking groups, order does not matter, so we need to divide:

(ABC is the same team as ACB, etc)

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12
Q

How many different ways are there to create a code, using 3 DISTINCT digits, using only integers greater than 1 and less than 8?

A

Order Matters, so don’t divide:

How many integers are greater than 1 and less than 8?

2, 3, 4, 5, 6, 7 —> 6 digits

Slots method —> 3 slots —> _6_ * _5_ *_4_ = 120

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13
Q

How many different codes can be made from the letters A A A B C?

A

There are 5 letters, and order matters

If they were all different letters, it would be:

5 * 4 * 3 * 2 = 120

When there are repeats, we divide by the # of repeats, Factorial

So, since there are 3 A’s, it is 120 / 3! = 120 / (3*2) = 20

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14
Q

How many different codes can be made from the letters A A A B B C?

A

There are 6 letters, and order matters

If they were all different letters, it would be:

6 * 5 * 4 * 3 * 2 = 720

When there are repeats, we divide by the # of repeats, Factorial

So, since there are 3 A’s and 2 B’s we need to divide by 3! and by 2!

720 / ( 3! * 2!) = 720 / (3*2*2) = 60

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15
Q

The diagram shows the various paths along which a mouse can travel from point X, where it is released, to point Y, where it is rewarded with a food pellet. How many different paths from X to Y can the mouse take if it goes directly from X to Y without retracing any point along a path?

A

Permutations (we multiply the number of options for each slot/choice/fork)

There are 3 forks along the path: 2 choices for the first one, 2 for the second and 3 for the third.

Total # of ways is 2*2*3 = 12

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16
Q

There are 6 women and 3 men. How many different teams of 2 women and 2 men are possible?

A

Count the women pairs and men pairs separately, then multiply for the total.

Women: 6 * 5 / 2 = 15

Men: 3 * 2 / 2 = 3

15 * 3 = 45

17
Q

5 athletes are competing for gold, silver and bronze medals, plus 1 honorable mention. How many different ways can the awards be distributed?

A

This time, the order matters, so we don’t divide by “4!”

4 Slots (for gold, silver, bronze, honorable mention) –> 5*4*3*2 = 20*6 = 120

18
Q

There are 9 candidates for 4 positions on a committee. How many different committees are possible?

A

126

out of 9, pick 4

4 slots on top, then divide by 4!, because the order doesn’t matter.