Examples

Simplifying conditionals inclusion

Let . Simplify the following values:

Solution

• Definition of ‘conditional’:
• The problem assumes that . Therefore .
• Therefore, answer: .

• Definition of ‘conditional’:
• Since , we know .
• Therefore and answer .

• Definition of ‘conditional’:
• Since , we have .
• Therefore, answer .

• Definition of ‘conditional’:
• There is no way to simplify further.
• We could write if desired.

Coin flipping: at least 2 heads

Flip a fair coin 4 times and record the outcomes as sequences, like HHTH.

Let be the event that there are at least two heads, and the event that there is at least one heads.

First let’s calculate .

Define , the event that there were exactly 2 heads, and , the event of exactly 3, and the event of exactly 4. These events are exclusive, so:

Each term on the right can be calculated by counting:

Therefore, .

Now suppose we find out that “at least one heads definitely came up”. (Meaning that we know .) For example, our friend is running the experiment and tells us this fact about the outcome.

Now what is our estimate of likelihood of ?

The formula for conditioning gives:

Now . (Any outcome with at least two heads automatically has at least one heads.) We already found that . To compute we simply add the probability , which is , to get .

Therefore:

Multiplication: flip a coin, then roll dice

Flip a coin. If the outcome is heads, roll two dice and add the numbers. If the outcome is tails, roll a single die and take that number. What is the probability of getting a tails AND a number at least 3?

Solution

(1) This “two-stage” experiment lends itself to a solution using the multiplication rule for conditional probability.

Label the events of interest.

Let and be the events that the coin showed heads and tails, respectively.

Let be the events that the final number is , respectively.

The value we seek is .

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(2) Observe known (conditional) probabilities.

We know that and .

We know that , for example, or that .

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(3) Apply multiplication rule:

We know and can see by counting that .

Therefore .

Multiplication: draw two cards

Two cards are drawn from a standard deck (without replacement).

What is the probability that the first is a 3, and the second is a 4?

Solution

This “two-stage” experiment lends itself to a solution using the multiplication rule for conditional probability.

(1) Label events:

• Write for the event that the first card is a 3.
• Write for the event that the second card is a 4.

We seek . We will use the multiplication rule:

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(2) Compute probabilities:

We know . (Does not depend on the second draw.)

For the conditional probability, note that if the first is a 3, then there are four remaining 4s and 51 remaining cards. Therefore:

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(3) Apply multiplication rule:

Marble transferred, marble drawn

Setup:

• Bin 1 holds five red and four green marbles.
• Bin 2 holds four red and five green marbles.

Experiment:

• You take a random marble from Bin 1 and put it in Bin 2 and shake Bin 2.
• Then you draw a random marble from Bin 2 and look at it.

What is the probability that the marble you look at is red?

Solution

(1) Label events:

• Event : a red marble is transferred.
• Event : a green marble is transferred.
• Event : a red marble is drawn from Bin 2.
• Event : a green marble is drawn from Bin 2.

Answer will be .

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(2) Apply Division into Cases:

General formula:

Insert our labels, and and . Obtain:

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(3) Plug in data and compute:

• Know .
• Know .
• Know .
• Know .

Therefore: