Dependent Events

Since both cards must be green, this is an “and” question. Though the events are the same, the first card is not replaced, so they will have different probabilities. Calculate both probabilities then multiply.

First card
Bottom: 52 outcomes are possible
Top: 13 cards are green
Probability: \dfrac{13}{52} = \dfrac{1}{4}

Second card
Bottom: 51 outcomes are possible since the first card was not replaced
Top: 12 cards are green, since one green was drawn and not replaced
Probability: \dfrac{12}{51} = \dfrac{4}{17}

Probability of pulling 2 greens without replacement:

This is a dependent probability. Once the first marble is pulled, there is one less marble for the second pull.

First pull
Bottom: 10 marbles total
Top: 6 marbles are black
Probability: \dfrac{6}{10} = \dfrac{3}{5}

If the first marble was black, there will now only be 5 black marbles left, and 9 total marbles.

Second pull
Bottom: 9 marbles left
Top: 5 marbles are black
Probability: \dfrac{5}{9}

Total probability:

Thinking through this problem logically will get you to the right answer. It is a dependent probability problem, because after each throw the number of sides the die can show decreases by 1. Remember to think bottom to top for each step.

	First throw: None of the sides are marked, so the die can land any side up and it will work. The probability is \dfrac{6}{6} = 1. The top side gets marked.
	Second throw: Only one side is marked, so the die can have any of the other 5 unmarked sides on top. The probability is \dfrac{5}{6}. The top gets marked.
	Third throw: Two sides are marked, so the die can have any of the 4 unmarked sides on top. The probability is \dfrac{4}{6} = \dfrac{2}{3}. The top gets marked.
	Fourth throw: Three sides are marked, so there are 3 unmarked sides. The probability is \dfrac{3}{6} = \dfrac{1}{2}. The top gets marked.
	Fifth throw: Four sides are marked, so there are only 2 unmarked. The probability of is \dfrac{2}{6} = \dfrac{1}{3}. The top gets marked.
	Sixth throw: Five sides are marked, so there is only one unmarked side. The probability is \dfrac{1}{6}.

The probability is the product of the probabilities from each throw.

\dfrac{6}{6} × \dfrac{5}{6} × \dfrac{2}{3} × \dfrac{1}{2} × \dfrac{1}{3} × \dfrac{1}{6}

= \dfrac{5}{6 × 3 × 3 × 6}\\[3ex]

Simplify the fraction, but wait before you calculate the denominator.

Notice that only two of the answers, (B) and (E), have 5 as the numerator. So either (B) or (E) is the answer.

Do a quick estimation of the denominator:

6 × 3 × 3 × 6 ≈ 6 × 10 × 6 = 360.

The denominator for (B) is 324 and for (E) is 216. Since 324 is closer to the estimate of 360, it is the best choice.

The correct answer is choice (B).

This question has both “or” and “and” in the same problem: independent and dependent.

For the “or” part, there are two desired scenarios that could happen: H H or T T.

But within each scenario, there are two scenarios, and both must happen (both must be heads or both must be tails).

Both Heads: This means heads on the first toss and heads on the second toss. The probability of each toss coming up heads is
\dfrac{1}{2} × \dfrac{1}{2} = \dfrac{1}{4}.

Both Tails: This has the same probability as heads:
\dfrac{1}{2} × \dfrac{1}{2} = \dfrac{1}{4}

Since these are independent events, add the probabilities.

Both Heads or Both Tails: 
\dfrac{1}{4} + \dfrac{1}{4} = \dfrac{2}{4} = \dfrac{1}{2}

The correct answer is choice (D).

There are just 2 possible scenarios.

1) A blue marble is pulled and put into the second bag and then a blue marble is pulled from the second bag.

2) A red marble is pulled and put into the second bag and then a blue marble is pulled from the second bag.

This question has both “or” and “and” in the same problem: independent and dependent.

For the “or” part, there are two desired scenarios that could happen: blue moved and blue pulled, or red moved and blue pulled.

But within each scenario, there are two scenarios, and both must happen (both blue moved and blue pulled, or both red moved and blue pulled).

Calculate the probability of each scenario and add them together.

The probability of the first scenario: The probability that a blue marble is pulled from the first basket is \dfrac{4}{9} (4 blue, 5 red). The probability that a blue marble is then pulled from the second basket is now \dfrac{4}{8} because there are now 4 blue marbles out of 8 total marbles. Multiply these dependent probabilities because it is an “and” scenario: \dfrac{4}{9} × \dfrac{4}{8} = \dfrac{16}{72}. Leave this as an unsimplified fraction until addition.

The probability of the second scenario: The probability that a red marble is taken from the first bag is \dfrac{5}{9}. The probability that a blue marble is then taken from the second bag = \dfrac{5}{9} × \dfrac{3}{8} = \dfrac{15}{72}.

The probability of EITHER the first OR the second scenarios: 
\dfrac{16}{72} + \dfrac{15}{72} = \dfrac{31}{72}.

The correct answer is choice (C).

Start by thinking about all the scenarios that could yield a sum of 7.

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

There are 6 pairs that sum to 7.

Method 1    Using probabilities

Each pair has the probability of \dfrac{1}{6} × \dfrac{1}{6} = \dfrac{1}{36}

These are 6 independent events, so “add” the probabilities by multiplying by 6.

6 × \dfrac{1}{36} = \dfrac{1}{6}

Method 2    Using outcomes

There are a total of 36 possible pairs of dice. You could make a list of them all (see the end of this section), or use combinations (covered in the previous chapter) to discover this.

There are 6 pairs that sum to 7.

So the probability is: