Probability and conditional probability

Almost every one of these questions says “selected at random.” The last line usually reads like one of these:

• “If one of these students is selected at random, what is the probability of selecting a student who plays a sport?”
• The same question with “given that the student is a senior” added at the end
• “How many of these students play a sport?”, after the question has given you a probability

The counts come from a table, a list of numbers, or a description in a sentence or two. On a grid-inA question where you type your own answer instead of picking one of four.Read the lesson, the question ends with “(Express your answer as a decimal or fraction, not as a percent.)”

A probability is the number of picks that fit the description over the number of picks you could make. If 12 of the 30 songs on a playlist are covers, the probability of picking a cover is $\frac{12}{30}=\frac{2}{5}$. It always comes out from 0 to 1, so an answer of 12 is a count that never got divided.

There's nothing to graph in these questions. Choosing the top and bottom of the fraction happens on paper. After that, Desmos can do the arithmetic.

Here are the 80 tickets sold for a school play.

If one of these tickets is selected at random, what is the probability of selecting a child ticket, given that the ticket is for the matinee?

Read to the end before you touch the table. “Given that the ticket is for the matinee” means you're only picking from matinee tickets, so every evening ticket is out.

1. Keep only the matinee column. These 36 tickets are the only ones you can pick.

   $$20+16=36$$

2. In that column, count the child tickets.

   $$16$$

3. Put the matinee total on the bottom and the child count on top.

   $$\frac{16}{36}$$

4. Divide the top and bottom by 4.

   $$=\frac{4}{9}$$

The top number always comes from inside the bottom group. Here the 16 child matinee tickets sit inside the 36 matinee tickets, so the fraction makes sense.

When the last line changes, the bottom can change too. Here are four questions about the same table.

A question about a child matinee ticket (or a ticket that's “child and matinee”) points at one cell, 16. Nothing in it rules out any tickets, though, so you're still picking from all 80.

When the condition covers more than one row, add those rows to get the bottom. This table covers every pair of shoes in a store, and every price is a whole number of dollars. It gives each cell as a share of all the pairs instead of a count, so the grand total is 1.00. You work it the same way.

If one of these pairs is selected at random, which of the following is closest to the probability that it's on sale, given that it costs less than $60?

1. Keep the under-$25 row and the $25 to $59 row. A $60 pair doesn't count, so the last row is out. Add the two totals for the bottom.

   $$0.31+0.42=0.73$$

2. In those two rows, add the on-sale shares for the top.

   $$0.23+0.16=0.39$$

3. Divide, and round only at the end.

   $$\frac{0.39}{0.73}\approx0.53$$

You don't need to turn the shares into counts first.

Whether a $60 pair counts comes down to the wording of the condition.

A band labeled “40 to 60” adds nothing to “more than 60,” since even its top value, 60, isn't more than 60.

A condition can have a “not” in it. “Given that it doesn't cost $60 or more” keeps the same two rows. You can also get there by subtracting the $60-or-more row: $1.00-0.27=0.73$ on the bottom and $0.46-0.07=0.39$ on top. Make both subtractions, or the top still counts pairs you've ruled out.

Now try one with counts in place of shares.

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At a rope walk, Salim prepares strands of three materials to be twisted into ropes. He sorts each strand into exactly one length category. The table gives the numbers of strands by material and length, in meters. One of these strands will be selected at random.

What is the probability of selecting a hemp strand, given that its length is at least 20 meters?

1. A

   $\frac{11}{50}$

2. B

   $\frac{11}{32}$

3. C

   $\frac{6}{17}$

4. D

   $\frac{11}{15}$

Show the answer

• A

  $\frac{11}{50}$

  Using $\frac{22}{100}=\frac{11}{50}$ puts all $100$ strands in the denominatorThe bottom number of a fraction, like the 4 in 3/4.Read the lesson. The condition “at least 20 meters” limits the denominator to the $64$ strands that meet that condition.

• B

  $\frac{11}{32}$

  Correct answer

  The length condition includes the “20 to less than 40 meters” and “At least 40 meters” columns, which contain $30+34=64$ strands. Of these, $10+12=22$ are hemp, so the probabilityHow likely something is, written as a number from 0 to 1, like 1/2 for heads on a coin flip.Read the lesson is $\frac{22}{64}=\frac{11}{32}$.

• C

  $\frac{6}{17}$

  Using $\frac{12}{34}=\frac{6}{17}$ counts only the strands at least $40$ meters long in both the numeratorThe top number of a fraction, like the 3 in 3/4.Read the lesson and denominatorThe bottom number of a fraction, like the 4 in 3/4.Read the lesson. This leaves out the “20 to less than 40 meters” category, which also satisfies “at least 20 meters.”

• D

  $\frac{11}{15}$

  Using $\frac{22}{30}=\frac{11}{15}$ reverses the condition: it gives the probabilityHow likely something is, written as a number from 0 to 1, like 1/2 for heads on a coin flip.Read the lesson that a strand is at least $20$ meters long given that it is hemp. The stated condition restricts the selection to strands at least $20$ meters long.

Solution

“Given that its length is at least 20 meters” restricts the possible selections to the “20 to less than 40 meters” and “At least 40 meters” columns. Add their totals to get $30+34=64$ eligible strands.

In the hemp row, those columns contain $10+12=22$ strands. Divide the number of eligible hemp strands by the total number of eligible strands: $\frac{22}{64}=\frac{11}{32}$.

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Sometimes there's no table, just a description. Build a small table yourself before you write the fraction.

A bakery made 6 trays of 15 muffins and 4 trays of 12 scones. By noon it had sold 2 trays of muffins and 1 tray of scones. If one of the items left is selected at random, what is the probability that it's a scone?

1. Muffin trays left.

   $$6-2=4$$

2. Scone trays left.

   $$4-1=3$$

3. Muffins left: 4 trays of 15.

   $$4\times15=60$$

4. Scones left: 3 trays of 12.

   $$3\times12=36$$

1. Add the items left. That's the group you're picking from.

   $$60+36=96$$

2. The scones left go on top.

   $$\frac{36}{96}$$

3. Divide the top and bottom by 12.

   $$=\frac{3}{8}$$

To count a set of whole numbers, divide. Of the whole numbers from 1 to 90, $90\div3=30$ are multiples of 3, so the probability of picking one is $\frac{30}{90}=\frac{1}{3}$.

Watch for a total that's then split into parts. “A cooler holds 20 bottles of juice, 12 apple and 8 grape, plus 15 bottles of water” describes 35 bottles. The 12 and the 8 are the same 20 juice bottles, so adding 20, 12, 8 and 15 counts every juice bottle twice. The probability of an apple juice is $\frac{12}{35}$.

Slow down at “not,” “neither,” “other than” and “outside.” Each one changes which picks count. Of the 80 play tickets, 36 are for the matinee. For a ticket that's not for the matinee, take those 36 out and count what's left.

1. Take the 36 matinee tickets out of all 80.

   $$80-36=44$$

2. Put what's left over all 80.

   $$\frac{44}{80}$$

3. Divide the top and bottom by 4.

   $$=\frac{11}{20}$$

That's the same as $1-\frac{36}{80}$.

“Neither A nor B” means not A and not B, so both groups come out.

A drawer holds 50 socks: 9 red, 5 green, 21 black and the rest gray. If one sock is selected at random, what is the probability that it's neither black nor gray?

1. A sock that's neither black nor gray is red or green. Count those.

   $$9+5=14$$

2. Put them over all 50 socks.

   $$\frac{14}{50}$$

3. Divide the top and bottom by 2.

   $$=\frac{7}{25}$$

That's quicker than taking both colors out of 50, since you'd have to work out the gray count first.

Two “nots” can come together. This table sorts 70 jackets.

For a jacket that's not large and doesn't have a hood, turn each “not” into what's left. Not large means small, and no hood means the no-hood column. The only cell that's both is small with no hood, 27, so the probability is $\frac{27}{70}$. Adding all 38 small jackets to all 40 no-hood jackets gives 78, more jackets than there are, because the 27 got counted twice.

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Quentin packed $86$ instrument cases into a shipping crate. Each case contains either a sensor unit or a tool unit and uses exactly one type of protective packing. The table shows the distribution of the cases.

If one of these cases is selected at random, what is the probability of selecting a case that uses neither a foam sleeve nor molded pulp?

1. A

   $\frac{41}{86}$

2. B

   $\frac{45}{86}$

3. C

   $\frac{25}{86}$

4. D

   $\frac{69}{86}$

Show the answer

• A

  $\frac{41}{86}$

  This adds the $17$ foam-sleeve cases and $24$ molded-pulp cases, giving the probabilityHow likely something is, written as a number from 0 to 1, like 1/2 for heads on a coin flip.Read the lesson of selecting one of the excluded packing types. The question asks for the complement of that event.

• B

  $\frac{45}{86}$

  Correct answer

  The qualifying cases use air pillows or kraft paper, so their count is $20+25=45$. Out of $86$ total cases, the probabilityHow likely something is, written as a number from 0 to 1, like 1/2 for heads on a coin flip.Read the lesson is $\frac{45}{86}$.

• C

  $\frac{25}{86}$

  This counts only the $25$ kraft-paper cases. It omits the $20$ air-pillow cases, which also use neither a foam sleeve nor molded pulp.

• D

  $\frac{69}{86}$

  This removes only the $17$ foam-sleeve cases from the total. The $24$ molded-pulp cases must also be removed.

Solution

A case that uses neither a foam sleeve nor molded pulp must use air pillows or kraft paper. The table shows $20$ air-pillow cases and $25$ kraft-paper cases, so the number that satisfy the condition is

$20+25=45$. There are $86$ cases in all. Therefore, the probabilityHow likely something is, written as a number from 0 to 1, like 1/2 for heads on a coin flip.Read the lesson is

$\frac{45}{86}$. The fraction cannot be reduced because $45$ and $86$ have no common factor greater than $1$.

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Here's the ticket table again.

If one ticket is selected at random, what is the probability that it's a child ticket or a matinee ticket? “Or” counts a ticket that's either one, or both.

1. Add the child total and the matinee total.

   $$30+36=66$$

2. The 16 child matinee tickets are in both totals. Take them out once.

   $$66-16=50$$

3. Put it over all 80 tickets.

   $$\frac{50}{80}$$

4. Divide the top and bottom by 10.

   $$=\frac{5}{8}$$

To check, count the cells that fit: 20 adult matinee, 16 child matinee and 14 child evening make 50. As a grid-in, enter 5/8 or .625.

Written with $P(\ )$ for “the probability of,” the rule looks like this.

When nothing can be in both groups, there's nothing to subtract. A soccer team scored 1, 0, 2, 1, 3, 0, 1, 4, 2, 1, 0 and 1 goals in its 12 games. Tally each value: 0 shows up three times and 1 shows up five times. A game can't end with both 0 and 1 goals, so the probability of picking a game with 0 or 1 goals is $\frac{3+5}{12}=\frac{8}{12}=\frac{2}{3}$.

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Cyrus recorded the merchandise sales at a concert. Each sale was for exactly one type of item and occurred during exactly one of the three time periods shown.

If one of these sales is selected at random, what is the probability that the sale was for a poster or occurred before the show? (Express your answer as a decimal or fraction, not as a percent.)

Show the answer

Accepted answers: 11/18, .6111

Solution

There are 31 poster sales and 21 sales before the show. The 8 posters sold before the show appear in both counts, so subtract them once: $31+21-8=44$. Divide by all 72 sales: $\frac{44}{72}=\frac{11}{18}=0.6111\ldots$. A response of $\frac{13}{18}$ adds $31+21$ without subtracting the 8 sales counted in both groups. A response of $\frac{44}{31}$ uses the union count of 44 but divides by the 31 poster sales instead of all 72 sales. A response of $\frac{7}{18}$ finds the probabilityHow likely something is, written as a number from 0 to 1, like 1/2 for heads on a coin flip.Read the lesson that a sale was neither for a poster nor before the show.

Note that $11/18$ and $.6111$ are examples of ways to enter a correct answer.

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Some questions give you a probability and ask how many. The count is the probability times the total.

A ferry carried 600 vehicles, and each was a car, a van or a motorcycle. Of these, 30 were motorcycles. If one vehicle is selected at random, the probability that it's a van is 0.15. How many of the vehicles were cars?

1. Turn the van probability into a count: multiply by the total.

   $$0.15\times600=90$$

2. Add the vans and the motorcycles.

   $$90+30=120$$

3. Every vehicle is in one of the three groups, so the cars are the rest.

   $$600-120=480$$

Say 6 of the 15 mugs on a shelf are blue. The probability of picking a blue one is $\frac{6}{15}$, which reduces to $\frac{2}{5}$.

If Desmos shows a long decimal, the small round fraction button on the result row turns it into a fraction that may fit the box. The grid-in lesson covers entry rules in more detail.

Before you start each one, find the group you're picking from and write its total on the bottom. Then check that everything you count on top is inside that group.