Examples

Simplifying conditionals inclusion

Let $A\subset B$. Simplify the following values:

$$P[A|B],P[A|B^c],P[B|A],P[B|A^c]$$

Solution

$P[A|B]$

• Definition of ‘conditional’: $P[A|B]=\frac{P[A\cap B]}{P[B]}$
• The problem assumes that $A\subset B$. Therefore $A\cap B=A$.
• Therefore, answer: $\frac{P[A]}{P[B]}$.

$P[A|B^c]$

• Definition of ‘conditional’: $P[A|B^c]=\frac{P[A\cap B^c]}{P[B^c]}$
• Since $A\subset B$, we know $A\cap B^c=\varnothing$.
• Therefore $P[A\cap B^c]=0$ and answer $=0$.

$P[B|A]$

• Definition of ‘conditional’: $P[B|A]=\frac{P[B\cap A]}{P[A]}$
• Since $A\subset B$, we have $B\cap A=A$.
• Therefore, answer $=1$.

$P[B|A^c]$

• Definition of ‘conditional’: $P[B|A^c]=\frac{P[B\cap A^c]}{P[A^c]}$
• There is no way to simplify further.
• We could write $P[A^c]=1-P[A]$ if desired.

Coin flipping: at least 2 heads

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

Let $A_{\ge 2}$ be the event that there are at least two heads, and $A_{\ge 1}$ the event that there is at least one heads.

First let’s calculate $P[A_{\ge 2}]$.

Define $A_2$, the event that there were exactly 2 heads, and $A_3$, the event of exactly 3, and $A_4$ the event of exactly 4. These events are exclusive, so:

$$P[A_{\ge 2}]=P[A_2\cup A_3\cup A_4]\gg \gg P[A_2]+P[A_3]+P[A_4]$$

Each term on the right can be calculated by counting:

$$\begin{array}{cc}P[A_2]& =\frac{|A_2|}{2^4}\gg \gg \frac{\left(\genfrac{}{}{0px}{}{4}{2}\right)}{16}\gg \gg \frac{6}{16}\\ & \\ P[A_3]& =\frac{|A_3|}{2^4}\gg \gg \frac{\left(\genfrac{}{}{0px}{}{4}{3}\right)}{16}\gg \gg \frac{4}{16}\\ & \\ P[A_4]& =\frac{|A_4|}{2^4}\gg \gg \frac{\left(\genfrac{}{}{0px}{}{4}{4}\right)}{16}\gg \gg \frac{1}{16}\end{array}$$

Therefore, $P[A_{\ge 2}]=\frac{11}{16}$.

Now suppose we find out that “at least one heads definitely came up”. (Meaning that we know $A_{\ge 1}$.) For example, our friend is running the experiment and tells us this fact about the outcome.

Now what is our estimate of likelihood of $A_{\ge 2}$?

The formula for conditioning gives:

$$P[A_{\ge 2}|A_{\ge 1}]=\frac{P[A_{\ge 2}\cap A_{\ge 1}]}{P[A_{\ge 1}]}$$

Now $A_{\ge 2}\cap A_{\ge 1}=A_{\ge 2}$. (Any outcome with at least two heads automatically has at least one heads.) We already found that $P[A_{\ge 2}]=\frac{11}{16}$. To compute $P[A_{\ge 1}]$ we simply add the probability $P[A_1]$, which is $\frac{4}{16}$, to get $P[A_{\ge 1}]=\frac{15}{16}$.

Therefore:

$$P[A_{\ge 2}|A_{\ge 1}]=\frac{11/16}{15/16}\gg \gg \frac{11}{15}$$

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 $H$ and $T$ be the events that the coin showed heads and tails, respectively.

Let $A_1,\dots ,A_{12}$ be the events that the final number is $1,\dots ,12$, respectively.

The value we seek is $P[T{A}_{\ge 3}]$.

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(2) Observe known (conditional) probabilities.

We know that $P[H]=\frac{1}{2}$ and $P[T]=\frac{1}{2}$.

We know that $P[A_5|T]=\frac{1}{6}$, for example, or that $P[A_2|H]=\frac{1}{36}$.

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(3) Apply multiplication rule:

$$P[T{A}_{\ge 3}]=P[T]\cdot P[A_{\ge 3}|T]$$

We know $P[T]=\frac{1}{2}$ and can see by counting that $P[A_{\ge 3}|T]=\frac{2}{3}$.

Therefore $P[T{A}_{\ge 3}]=\frac{1}{3}$.

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

We seek $P[TF]$. We will use the multiplication rule:

$$P[TF]=P[T]\cdot P[F|T]$$

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(2) Compute probabilities:

We know $P[T]=\frac{4}{52}$. (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:

$$P[F|T]=\frac{4}{51}$$

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(3) Apply multiplication rule:

$$\begin{array}{c}P[TF]=P[T]\cdot P[F|T]\\ \\ \\ P[TF]=\frac{4}{52}\cdot \frac{4}{51}\gg \gg \frac{4}{13\cdot 51}\end{array}$$

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 $T_R$: a red marble is transferred.
• Event $T_G$: a green marble is transferred.
• Event $D_R$: a red marble is drawn from Bin 2.
• Event $D_G$: a green marble is drawn from Bin 2.

Answer will be $P[D_R]$.

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(2) Apply Division into Cases:

General formula: $P[B]=P[A]\cdot P[B|A]+P[A^c]\cdot P[B|A^c]$

Insert our labels, $A\to T_R$ and $B\to D_R$ and $A^c\to T_G$. Obtain:

$$P[D_R]=P[T_R]\cdot P[D_R|T_R]+P[T_G]\cdot P[D_R|T_G]$$

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(3) Plug in data and compute:

• Know $P[T_R]=5/9$.
• Know $P[T_G]=4/9$.
• Know $P[D_R|T_R]=5/10$.
• Know $P[D_R|T_G]=4/10$.

Therefore:

$$P[D_R]\gg \gg \frac{5}{9}\cdot \frac{5}{10}+\frac{4}{9}\cdot \frac{4}{10}\gg \gg \frac{41}{90}$$