Absolutely. This time I’m using actual Unicode mathematics throughout rather than LaTeX, and I’ve removed College Board and other website references from the article itself.
𝗦𝗔𝗧 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 & 𝗖𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻𝗮𝗹 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆
𝗧𝗵𝗲 𝗖𝗼𝗺𝗽𝗹𝗲𝘁𝗲 𝗚𝘂𝗶𝗱𝗲 𝘁𝗼 𝗦𝗮𝗺𝗽𝗹𝗲 𝗦𝗽𝗮𝗰𝗲𝘀, 𝗧𝗮𝗯𝗹𝗲𝘀, 𝗖𝗼𝘂𝗻𝘁𝗶𝗻𝗴 & 𝗧𝗿𝗶𝗰𝗸𝘆 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻𝘀
Probability questions can look much harder than they really are.
A question may talk about students, games, cards, machines, surveys, products or randomly selected objects. But underneath the story, the mathematics usually comes down to one central idea:
⭐ 𝗛𝗼𝘄 𝗺𝗮𝗻𝘆 𝗽𝗼𝘀𝘀𝗶𝗯𝗹𝗲 𝗼𝘂𝘁𝗰𝗼𝗺𝗲𝘀 𝗮𝗿𝗲 𝘁𝗵𝗲𝗿𝗲, 𝗮𝗻𝗱 𝗵𝗼𝘄 𝗺𝗮𝗻𝘆 𝗺𝗮𝘁𝗰𝗵 𝘄𝗵𝗮𝘁 𝗜 𝘄𝗮𝗻𝘁?
The biggest challenge is often not calculation.
It is identifying the correct group, especially when a question contains words such as:
𝗴𝗶𝘃𝗲𝗻 𝘁𝗵𝗮𝘁
𝗮𝗺𝗼𝗻𝗴
𝗼𝗳 𝘁𝗵𝗼𝘀𝗲 𝘄𝗵𝗼
𝗮𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲
𝘄𝗶𝘁𝗵𝗼𝘂𝘁 𝗿𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁
Master those phrases and many SAT probability questions become much easier.
① 𝗧𝗵𝗲 𝗕𝗮𝘀𝗶𝗰 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗙𝗼𝗿𝗺𝘂𝗹𝗮
When all outcomes are equally likely:
𝗣(𝗘) ＝ 𝗙𝗮𝘃𝗼𝗿𝗮𝗯𝗹𝗲 𝗢𝘂𝘁𝗰𝗼𝗺𝗲𝘀 ／ 𝗧𝗼𝘁𝗮𝗹 𝗢𝘂𝘁𝗰𝗼𝗺𝗲𝘀
𝗘𝘅𝗮𝗺𝗽𝗹𝗲
A box contains 8 red balls and 4 blue balls.
One ball is selected randomly.
Total balls:
８ ＋ ４ ＝ １２
Favorable outcomes for blue:
４
Therefore:
𝗣(𝗯𝗹𝘂𝗲) ＝ ４／１２ ＝ １／３
⭐ 𝗦𝗔𝗧 𝗦𝗲𝗰𝗿𝗲𝘁
Before calculating anything, ask:
“𝗪𝗵𝗮𝘁 𝗶𝘀 𝗺𝘆 𝗱𝗲𝗻𝗼𝗺𝗶𝗻𝗮𝘁𝗼𝗿?”
The wrong denominator is one of the easiest ways to lose a probability question.
② 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗖𝗮𝗻𝗻𝗼𝘁 𝗕𝗲 𝗟𝗲𝘀𝘀 𝗧𝗵𝗮𝗻 𝟬 𝗼𝗿 𝗚𝗿𝗲𝗮𝘁𝗲𝗿 𝗧𝗵𝗮𝗻 𝟭
Every probability satisfies:
０ ≤ 𝗣(𝗘) ≤ １
For example:
１／４ ＝ ０．２５ ＝ ２５％
A probability of:
０
means the event is impossible.
A probability of:
１
means the event is certain.
Therefore, an answer such as:
１．２
cannot be a probability.
③ 𝗧𝗵𝗲 𝗖𝗼𝗺𝗽𝗹𝗲𝗺𝗲𝗻𝘁 𝗧𝗿𝗶𝗰𝗸
The complement of an event means that the event does not happen.
𝗣(𝗻𝗼𝘁 𝗘) ＝ １ − 𝗣(𝗘)
This is particularly useful when you see:
𝗮𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲
𝗻𝗼𝗻𝗲
𝗻𝗼𝘁
𝗻𝗲𝘃𝗲𝗿
𝗱𝗼𝗲𝘀 𝗻𝗼𝘁
𝗘𝘅𝗮𝗺𝗽𝗹𝗲
A machine produces a defective item with probability ０．０８.
What is the probability that an item is not defective?
１ − ０．０８ ＝ ０．９２
Therefore:
𝗔𝗻𝘀𝘄𝗲𝗿 ＝ ９２％
④ 𝗧𝗵𝗲 “𝗔𝗡𝗗” 𝗥𝘂𝗹𝗲
When two independent events must both occur:
𝗣(𝗔 ∩ 𝗕) ＝ 𝗣(𝗔) × 𝗣(𝗕)
The symbol:
∩
means intersection, or the outcome where both events occur.
𝗘𝘅𝗮𝗺𝗽𝗹𝗲
A fair coin is tossed twice.
What is the probability of getting heads both times?
𝗣(𝗛) ＝ １／２
Therefore:
𝗣(𝗛 ∩ 𝗛) ＝ １／２ × １／２
＝ １／４
⭐ 𝗠𝗲𝗺𝗼𝗿𝘆 𝗧𝗿𝗶𝗰𝗸
𝗔𝗡𝗗 → 𝗧𝗵𝗶𝗻𝗸 𝗠𝗨𝗟𝗧𝗜𝗣𝗟𝗬
But remember: multiplication assumes the appropriate independence or conditional structure.
⑤ 𝗧𝗵𝗲 “𝗢𝗥” 𝗥𝘂𝗹𝗲
The symbol:
∪
means union, or an outcome belonging to at least one of the events.
If two events cannot overlap:
𝗣(𝗔 ∪ 𝗕) ＝ 𝗣(𝗔) ＋ 𝗣(𝗕)
𝗘𝘅𝗮𝗺𝗽𝗹𝗲
A number from １ through １０ is selected.
What is the probability of selecting ２ or ９?
There are １０ possible numbers.
Favorable outcomes:
２
Therefore:
𝗣(２ 𝗼𝗿 ９) ＝ ２／１０ ＝ １／５
⑥ 𝗧𝗵𝗲 𝗢𝘃𝗲𝗿𝗹𝗮𝗽 𝗧𝗿𝗮𝗽
Sometimes two events overlap.
Then simply adding their probabilities counts the shared outcomes twice.
The general rule is:
𝗣(𝗔 ∪ 𝗕) ＝ 𝗣(𝗔) ＋ 𝗣(𝗕) − 𝗣(𝗔 ∩ 𝗕)
𝗘𝘅𝗮𝗺𝗽𝗹𝗲
In a group of students:
• ２８ play basketball
• ２２ play soccer
• ９ play both
How many play at least one of the two sports?
Start with:
２８ ＋ ２２ ＝ ５０
The ９ students who play both were counted twice.
Subtract them:
５０ − ９ ＝ ４１
So:
４１ students
play at least one sport.
⭐ 𝗥𝗲𝗺𝗲𝗺𝗯𝗲𝗿
𝗢𝗥 → 𝗔𝗗𝗗
If there is overlap:
𝗔𝗗𝗗 → 𝗦𝗨𝗕𝗧𝗥𝗔𝗖𝗧 𝗧𝗛𝗘 𝗢𝗩𝗘𝗥𝗟𝗔𝗣
⑦ 𝗜𝗻𝗱𝗲𝗽𝗲𝗻𝗱𝗲𝗻𝘁 𝗘𝘃𝗲𝗻𝘁𝘀
Two events are independent when the occurrence of one does not change the probability of the other.
For independent events:
𝗣(𝗔 ∩ 𝗕) ＝ 𝗣(𝗔) × 𝗣(𝗕)
𝗘𝘅𝗮𝗺𝗽𝗹𝗲
A die is rolled and a coin is flipped.
The die result does not affect the coin result.
Therefore:
𝗣(６ 𝗮𝗻𝗱 𝗵𝗲𝗮𝗱𝘀)
＝ １／６ × １／２
＝ １／１２
⑧ 𝗗𝗲𝗽𝗲𝗻𝗱𝗲𝗻𝘁 𝗘𝘃𝗲𝗻𝘁𝘀
Sometimes the first event changes the probability of the second.
This commonly happens when objects are selected without replacement.
𝗘𝘅𝗮𝗺𝗽𝗹𝗲
A bag contains:
５ red balls
７ blue balls
Two balls are selected without replacement.
Probability of getting two red balls:
First red:
５／１２
After removing one red ball:
４／１１
Therefore:
５／１２ × ４／１１
＝ ２０／１３２
＝ ５／３３
⭐ 𝗖𝗿𝗶𝘁𝗶𝗰𝗮𝗹 𝗣𝗼𝗶𝗻𝘁
𝗪𝗶𝘁𝗵 𝗿𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁 → 𝘁𝗵𝗲 𝗴𝗿𝗼𝘂𝗽 𝗿𝗲𝘁𝘂𝗿𝗻𝘀 𝘁𝗼 𝗶𝘁𝘀 𝗼𝗿𝗶𝗴𝗶𝗻𝗮𝗹 𝘀𝘁𝗮𝘁𝗲
𝗪𝗶𝘁𝗵𝗼𝘂𝘁 𝗿𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁 → 𝘁𝗵𝗲 𝗴𝗿𝗼𝘂𝗽 𝗰𝗵𝗮𝗻𝗴𝗲𝘀
That difference can completely change the answer.
⑨ 𝗖𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻𝗮𝗹 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆
Conditional probability is the probability of one event after another condition has already been established.
It is written:
𝗣(𝗔｜𝗕)
Read it as:
“the probability of A given B.”
The fundamental formula is:
𝗣(𝗔｜𝗕) ＝ 𝗣(𝗔 ∩ 𝗕) ／ 𝗣(𝗕)
But for SAT questions, there is an even more useful way to remember it:
⭐ 𝗧𝗛𝗘 𝗖𝗢𝗡𝗗𝗜𝗧𝗜𝗢𝗡 𝗖𝗛𝗔𝗡𝗚𝗘𝗦 𝗧𝗛𝗘 𝗗𝗘𝗡𝗢𝗠𝗜𝗡𝗔𝗧𝗢𝗥.
⑩ 𝗧𝗵𝗲 “𝗡𝗲𝘄 𝗧𝗼𝘁𝗮𝗹” 𝗧𝗿𝗶𝗰𝗸
Suppose a school has １００ students.
• ６０ are juniors.
• ２４ juniors play tennis.
A student is selected from the juniors.
What is the probability that the student plays tennis?
You are no longer choosing from all １００ students.
Your new group is:
６０ juniors
Of those:
２４ play tennis
Therefore:
𝗣(𝘁𝗲𝗻𝗻𝗶𝘀｜𝗷𝘂𝗻𝗶𝗼𝗿) ＝ ２４／６０
＝ ２／５
＝ ４０％
🚨 𝗧𝗵𝗲 𝘁𝗿𝗮𝗽
Do not calculate:
２４／１００
That answers a different question.
The condition “junior” has already reduced your sample space.
⑪ 𝗧𝘄𝗼-𝗪𝗮𝘆 𝗧𝗮𝗯𝗹𝗲𝘀
Two-way tables are extremely useful for conditional probability.
Consider:
| 𝗦𝗽𝗼𝗿𝘁𝘀 | 𝗡𝗼 𝗦𝗽𝗼𝗿𝘁𝘀 | 𝗧𝗼𝘁𝗮𝗹 | |
|---|---|---|---|
| 𝗝𝘂𝗻𝗶𝗼𝗿𝘀 | ２４ | ３６ | ６０ |
| 𝗦𝗲𝗻𝗶𝗼𝗿𝘀 | １０ | ３０ | ４０ |
| 𝗧𝗼𝘁𝗮𝗹 | ３４ | ６６ | １００ |
Question:
Given that a student is a junior, what is the probability that the student plays sports?
The condition is:
𝗝𝘂𝗻𝗶𝗼𝗿
Therefore, use the junior total:
６０
Favorable students:
２４
So:
𝗣(𝘀𝗽𝗼𝗿𝘁𝘀｜𝗷𝘂𝗻𝗶𝗼𝗿) ＝ ２４／６０
＝ ２／５
＝ ４０％
🔑 𝗧𝗵𝗲 𝟯-𝗦𝗲𝗰𝗼𝗻𝗱 𝗧𝗿𝗶𝗰𝗸
When you see:
“𝗴𝗶𝘃𝗲𝗻 𝘁𝗵𝗮𝘁…”
immediately ask:
“𝗪𝗵𝗮𝘁 𝗴𝗿𝗼𝘂𝗽 𝗮𝗺 𝗜 𝗻𝗼𝘄 𝗹𝗼𝗼𝗸𝗶𝗻𝗴 𝗮𝘁?”
That group usually supplies the denominator.
⑫ 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗪𝗶𝘁𝗵 𝗣𝗲𝗿𝗰𝗲𝗻𝘁𝗮𝗴𝗲𝘀
Suppose:
７０％ of students own a laptop.
Among laptop owners:
４０％ also own a tablet.
What percentage of all students own both?
Translate:
𝗣(𝗟) ＝ ０．７０
and:
𝗣(𝗧｜𝗟) ＝ ０．４０
Therefore:
𝗣(𝗟 ∩ 𝗧) ＝ ０．７０ × ０．４０
＝ ０．２８
Therefore:
𝗔𝗻𝘀𝘄𝗲𝗿 ＝ ２８％
⑬ 𝗧𝗵𝗲 “𝗔𝘁 𝗟𝗲𝗮𝘀𝘁 𝗢𝗻𝗲” 𝗦𝗵𝗼𝗿𝘁𝗰𝘂𝘁
“At least one” means:
one or more.
Instead of calculating every possibility separately, calculate the probability of none.
Then subtract from １.
Suppose an event has probability:
０．２０
on each independent trial.
Over two trials:
Probability of failure each time:
１ − ０．２０ ＝ ０．８０
Probability of no success twice:
０．８０ × ０．８０ ＝ ０．６４
Therefore:
𝗣(𝗮𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲) ＝ １ − ０．６４
＝ ０．３６
＝ ３６％
⭐ 𝗠𝗲𝗺𝗼𝗿𝘆 𝗧𝗿𝗶𝗰𝗸
𝗔𝗧 𝗟𝗘𝗔𝗦𝗧 𝗢𝗡𝗘 ＝ １ − 𝗡𝗢𝗡𝗘
⑭ 𝗧𝗿𝗲𝗲 𝗗𝗶𝗮𝗴𝗿𝗮𝗺 𝗧𝗵𝗶𝗻𝗸𝗶𝗻𝗴
Tree diagrams are useful when a probability problem happens in stages.
A box contains:
３ green balls
２ yellow balls
Two balls are selected without replacement.
First green:
３／５
After one green is removed:
２ green + ２ yellow = ４ balls
Second green:
２／４ ＝ １／２
Therefore:
𝗣(𝗴𝗿𝗲𝗲𝗻 𝗮𝗻𝗱 𝗴𝗿𝗲𝗲𝗻)
＝ ３／５ × １／２
＝ ３／１０
🧠 𝗧𝗿𝗲𝗲 𝗗𝗶𝗮𝗴𝗿𝗮𝗺 𝗥𝘂𝗹𝗲
𝗠𝘂𝗹𝘁𝗶𝗽𝗹𝘆 𝗮𝗹𝗼𝗻𝗴 𝗮 𝗯𝗿𝗮𝗻𝗰𝗵.
𝗔𝗱𝗱 𝘀𝗲𝗽𝗮𝗿𝗮𝘁𝗲 𝗯𝗿𝗮𝗻𝗰𝗵𝗲𝘀 𝘁𝗵𝗮𝘁 𝗽𝗿𝗼𝗱𝘂𝗰𝗲 𝘁𝗵𝗲 𝗱𝗲𝘀𝗶𝗿𝗲𝗱 𝗿𝗲𝘀𝘂𝗹𝘁.
⑮ 𝗖𝗼𝘂𝗻𝘁𝗶𝗻𝗴 𝗣𝗼𝘀𝘀𝗶𝗯𝗹𝗲 𝗢𝘂𝘁𝗰𝗼𝗺𝗲𝘀
Sometimes the fastest probability method is simply counting.
Suppose a code contains:
• one digit from ０–９
• one letter from Ａ–Ｚ
There are:
１０ × ２６ ＝ ２６０
possible codes.
If one particular code is selected:
𝗣(𝘁𝗵𝗮𝘁 𝗰𝗼𝗱𝗲) ＝ １／２６０
⭐ 𝗖𝗼𝘂𝗻𝘁𝗶𝗻𝗴 𝗣𝗿𝗶𝗻𝗰𝗶𝗽𝗹𝗲
If one stage has m possibilities and another has n possibilities:
𝗧𝗼𝘁𝗮𝗹 𝗽𝗼𝘀𝘀𝗶𝗯𝗶𝗹𝗶𝘁𝗶𝗲𝘀 ＝ m × n
⑯ 𝗘𝘅𝗽𝗲𝗿𝗶𝗺𝗲𝗻𝘁𝗮𝗹 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆
Probability can also be estimated from actual results.
Suppose a spinner is used:
２００ times
It lands on blue:
５８ times
Experimental probability:
５８／２００ ＝ ０．２９
Therefore:
𝗘𝘅𝗽𝗲𝗿𝗶𝗺𝗲𝗻𝘁𝗮𝗹 𝗽𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 ＝ ２９％
𝗗𝗼𝗻'𝘁 𝗖𝗼𝗻𝗳𝘂𝘀𝗲
𝗧𝗵𝗲𝗼𝗿𝗲𝘁𝗶𝗰𝗮𝗹 𝗽𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆
comes from the mathematical structure.
𝗘𝘅𝗽𝗲𝗿𝗶𝗺𝗲𝗻𝘁𝗮𝗹 𝗽𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆
comes from observed results.
⑰ 𝗘𝘅𝗽𝗲𝗰𝘁𝗲𝗱 𝗩𝗮𝗹𝘂𝗲
Expected value describes the long-run average outcome.
Suppose a game gives:
• ₹１００ with probability ０．２０
• ₹０ with probability ０．８０
Then:
𝗘𝘅𝗽𝗲𝗰𝘁𝗲𝗱 𝗩𝗮𝗹𝘂𝗲
＝ １００ × ０．２０ ＋ ０ × ０．８０
＝ ２０
Therefore:
𝗘𝘅𝗽𝗲𝗰𝘁𝗲𝗱 𝘃𝗮𝗹𝘂𝗲 ＝ ₹２０
This does not mean every player receives ₹２０.
It means that over many plays, the average outcome approaches ₹２０ per play.
⑱ 𝗔 𝗛𝗮𝗿𝗱𝗲𝗿 𝗖𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻𝗮𝗹 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗣𝗿𝗼𝗯𝗹𝗲𝗺
A survey contains ５００ people.
• ３００ use App A
• ２５０ use App B
• １５０ use both
A person is selected from the people who use App B.
What is the probability that the person also uses App A?
The phrase:
“from the people who use App B”
changes the sample space.
The new total is:
２５０
The favorable group is:
１５０
Therefore:
𝗣(𝗔｜𝗕) ＝ １５０／２５０
＝ ３／５
＝ ６０％
❌ 𝗧𝗵𝗲 𝗧𝗿𝗮𝗽
You might calculate:
１５０／５００ ＝ ３０％
But that is the probability of selecting someone who uses both apps from the entire population.
The question does not ask that.
It asks for the probability among App B users.
⑲ 𝗦𝗔𝗧 𝗪𝗼𝗿𝗱𝘀 𝗧𝗵𝗮𝘁 𝗖𝗵𝗮𝗻𝗴𝗲 𝘁𝗵𝗲 𝗠𝗮𝘁𝗵
“𝗚𝗶𝘃𝗲𝗻 𝘁𝗵𝗮𝘁…”
Think:
𝗖𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻𝗮𝗹 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆
“𝗔𝗻𝗱”
Think:
𝗕𝗼𝘁𝗵 𝗺𝘂𝘀𝘁 𝗵𝗮𝗽𝗽𝗲𝗻
Often:
𝗠𝘂𝗹𝘁𝗶𝗽𝗹𝘆
“𝗢𝗿”
Think:
𝗘𝗶𝘁𝗵𝗲𝗿 𝗰𝗮𝗻 𝘀𝗮𝘁𝗶𝘀𝗳𝘆 𝘁𝗵𝗲 𝗰𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻
Usually:
𝗔𝗱𝗱
But check for overlap.
“𝗔𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲”
Think:
１ − 𝗻𝗼𝗻𝗲
“𝗪𝗶𝘁𝗵𝗼𝘂𝘁 𝗿𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁”
Think:
𝗗𝗲𝗽𝗲𝗻𝗱𝗲𝗻𝘁 𝗽𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝗶𝗲𝘀
“𝗔𝗺𝗼𝗻𝗴…”
Think:
𝗡𝗲𝘄 𝗱𝗲𝗻𝗼𝗺𝗶𝗻𝗮𝘁𝗼𝗿
⑳ 𝗧𝗵𝗲 𝗕𝗶𝗴𝗴𝗲𝘀𝘁 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗧𝗿𝗮𝗽𝘀
❌ 𝗧𝗿𝗮𝗽 𝟭 — 𝗪𝗿𝗼𝗻𝗴 𝗱𝗲𝗻𝗼𝗺𝗶𝗻𝗮𝘁𝗼𝗿
If the question gives a condition, the denominator may no longer be the grand total.
❌ 𝗧𝗿𝗮𝗽 𝟮 — 𝗙𝗼𝗿𝗴𝗲𝘁𝘁𝗶𝗻𝗴 𝗼𝘃𝗲𝗿𝗹𝗮𝗽
If two groups share members:
𝗔 ＋ 𝗕
may count some people twice.
❌ 𝗧𝗿𝗮𝗽 𝟯 — 𝗜𝗴𝗻𝗼𝗿𝗶𝗻𝗴 “𝘄𝗶𝘁𝗵𝗼𝘂𝘁 𝗿𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁”
The second probability may be different.
❌ 𝗧𝗿𝗮𝗽 𝟰 — 𝗖𝗼𝗻𝗳𝘂𝘀𝗶𝗻𝗴 “𝗮𝘁 𝗹𝗲𝗮𝘀𝘁” 𝘄𝗶𝘁𝗵 “𝗲𝘅𝗮𝗰𝘁𝗹𝘆”
𝗔𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲
means one or more.
𝗘𝘅𝗮𝗰𝘁𝗹𝘆 𝗼𝗻𝗲
means one and only one.
❌ 𝗧𝗿𝗮𝗽 𝟱 — 𝗟𝗲𝘁𝘁𝗶𝗻𝗴 𝘁𝗵𝗲 𝗰𝗮𝗹𝗰𝘂𝗹𝗮𝘁𝗼𝗿 𝗱𝗲𝗰𝗶𝗱𝗲 𝘁𝗵𝗲 𝗺𝗮𝘁𝗵
A calculator can calculate the wrong expression perfectly.
𝗥𝗲𝗮𝘀𝗼𝗻 𝗳𝗶𝗿𝘀𝘁.
𝗖𝗮𝗹𝗰𝘂𝗹𝗮𝘁𝗲 𝘀𝗲𝗰𝗼𝗻𝗱.
㉑ 𝗢𝗿𝗶𝗴𝗶𝗻𝗮𝗹 𝗣𝗿𝗮𝗰𝘁𝗶𝗰𝗲 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻𝘀
🟢 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻 𝟭
A box contains ９ white counters and ６ black counters.
One counter is selected randomly.
What is the probability of selecting a black counter?
Total:
９ ＋ ６ ＝ １５
Therefore:
𝗣(𝗯𝗹𝗮𝗰𝗸) ＝ ６／１５
＝ ２／５
✅ 𝗔𝗻𝘀𝘄𝗲𝗿: ２／５
🟢 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻 𝟮
A fair die is rolled twice.
What is the probability that both results are even?
Even results:
２，４，６
Therefore:
𝗣(𝗲𝘃𝗲𝗻) ＝ ３／６ ＝ １／２
Both rolls even:
１／２ × １／２
＝ １／４
✅ 𝗔𝗻𝘀𝘄𝗲𝗿: １／４
🟡 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻 𝟯
A club has ４０ members.
２５ study mathematics.
１０ of those mathematics students also study physics.
If a mathematics student is selected randomly, what is the probability that the student studies physics?
The condition says:
𝗺𝗮𝘁𝗵𝗲𝗺𝗮𝘁𝗶𝗰𝘀 𝘀𝘁𝘂𝗱𝗲𝗻𝘁
Therefore:
𝗗𝗲𝗻𝗼𝗺𝗶𝗻𝗮𝘁𝗼𝗿 ＝ ２５
Favorable:
１０
Therefore:
𝗣(𝗽𝗵𝘆𝘀𝗶𝗰𝘀｜𝗺𝗮𝘁𝗵) ＝ １０／２５
＝ ２／５
✅ 𝗔𝗻𝘀𝘄𝗲𝗿: ２／５
🟡 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻 𝟰
A box contains ５ red counters and ７ blue counters.
Two counters are selected without replacement.
What is the probability that both are red?
First red:
５／１２
Second red:
４／１１
Therefore:
５／１２ × ４／１１
＝ ２０／１３２
＝ ５／３３
✅ 𝗔𝗻𝘀𝘄𝗲𝗿: ５／３３
🟠 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻 𝟱
A survey of １２０ students shows:
• ７０ take mathematics
• ５５ take physics
• ３５ take both
A mathematics student is selected randomly.
What is the probability that the student also takes physics?
The condition is:
𝗺𝗮𝘁𝗵𝗲𝗺𝗮𝘁𝗶𝗰𝘀
So:
𝗗𝗲𝗻𝗼𝗺𝗶𝗻𝗮𝘁𝗼𝗿 ＝ ７０
Both:
３５
Therefore:
𝗣(𝗽𝗵𝘆𝘀𝗶𝗰𝘀｜𝗺𝗮𝘁𝗵) ＝ ３５／７０
＝ １／２
✅ 𝗔𝗻𝘀𝘄𝗲𝗿: １／２
🔴 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻 𝟲
An event has a probability of ０．１５ on each independent trial.
What is the probability that it occurs at least once in three trials?
Probability it does not occur:
１ − ０．１５ ＝ ０．８５
Probability it never occurs:
０．８５ × ０．８５ × ０．８５
＝ ０．６１４１２５
Therefore:
𝗣(𝗮𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲)
＝ １ − ０．６１４１２５
＝ ０．３８５８７５
Therefore:
✅ 𝗔𝗻𝘀𝘄𝗲𝗿: ３８．５８７５％
㉒ 𝗧𝗵𝗲 𝗨𝗹𝘁𝗶𝗺𝗮𝘁𝗲 𝗦𝗔𝗧 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗖𝗵𝗲𝗮𝘁 𝗦𝗵𝗲𝗲𝘁
𝗕𝗮𝘀𝗶𝗰 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆
𝗣(𝗘) ＝ 𝗙𝗮𝘃𝗼𝗿𝗮𝗯𝗹𝗲 ／ 𝗧𝗼𝘁𝗮𝗹
𝗖𝗼𝗺𝗽𝗹𝗲𝗺𝗲𝗻𝘁
𝗣(𝗻𝗼𝘁 𝗘) ＝ １ − 𝗣(𝗘)
𝗜𝗻𝘁𝗲𝗿𝘀𝗲𝗰𝘁𝗶𝗼𝗻
𝗣(𝗔 ∩ 𝗕)
means A and B.
𝗨𝗻𝗶𝗼𝗻
𝗣(𝗔 ∪ 𝗕)
means A or B.
𝗖𝗼𝗻𝗱𝗶𝘁𝗶𝗼𝗻𝗮𝗹 𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆
𝗣(𝗔｜𝗕) ＝ 𝗣(𝗔 ∩ 𝗕) ／ 𝗣(𝗕)
𝗜𝗻𝗱𝗲𝗽𝗲𝗻𝗱𝗲𝗻𝘁 𝗘𝘃𝗲𝗻𝘁𝘀
𝗣(𝗔 ∩ 𝗕) ＝ 𝗣(𝗔) × 𝗣(𝗕)
𝗔𝘁 𝗟𝗲𝗮𝘀𝘁 𝗢𝗻𝗲
𝗣(𝗮𝘁 𝗹𝗲𝗮𝘀𝘁 𝗼𝗻𝗲) ＝ １ − 𝗣(𝗻𝗼𝗻𝗲)
𝗪𝗶𝘁𝗵𝗼𝘂𝘁 𝗥𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁
𝗧𝗵𝗲 𝗽𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗺𝗮𝘆 𝗰𝗵𝗮𝗻𝗴𝗲.
𝗪𝗶𝘁𝗵 𝗥𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁
𝗧𝗵𝗲 𝗼𝗿𝗶𝗴𝗶𝗻𝗮𝗹 𝗰𝗼𝗺𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 𝗶𝘀 𝗿𝗲𝘀𝘁𝗼𝗿𝗲𝗱.
🏆 𝗧𝗵𝗲 𝗢𝗻𝗲 𝗜𝗱𝗲𝗮 𝗬𝗼𝘂 𝗠𝘂𝘀𝘁 𝗥𝗲𝗺𝗲𝗺𝗯𝗲𝗿
The most important probability skill is not memorizing a dozen formulas.
It is identifying the correct sample space.
If the question asks about everyone:
𝗨𝘀𝗲 𝘁𝗵𝗲 𝗳𝘂𝗹𝗹 𝗴𝗿𝗼𝘂𝗽.
If it asks about students who belong to a particular category:
𝗨𝘀𝗲 𝘁𝗵𝗮𝘁 𝗴𝗿𝗼𝘂𝗽 𝗮𝘀 𝘆𝗼𝘂𝗿 𝗻𝗲𝘄 𝘀𝗮𝗺𝗽𝗹𝗲 𝘀𝗽𝗮𝗰𝗲.
If you see:
“𝗴𝗶𝘃𝗲𝗻 𝘁𝗵𝗮𝘁…”
stop and identify the condition.
Then think:
𝗖𝗢𝗡𝗗𝗜𝗧𝗜𝗢𝗡 → 𝗡𝗘𝗪 𝗚𝗥𝗢𝗨𝗣 → 𝗡𝗘𝗪 𝗗𝗘𝗡𝗢𝗠𝗜𝗡𝗔𝗧𝗢𝗥
That single habit can turn a confusing probability problem into a one-line calculation.
⭐ 𝗙𝗜𝗡𝗔𝗟 𝗠𝗘𝗠𝗢𝗥𝗬 𝗧𝗥𝗜𝗖𝗞
𝗣𝗿𝗼𝗯𝗮𝗯𝗶𝗹𝗶𝘁𝘆 𝗶𝘀 𝗻𝗼𝘁 𝗮𝗯𝗼𝘂𝘁 𝗴𝘂𝗲𝘀𝘀𝗶𝗻𝗴.
𝗜𝘁 𝗶𝘀 𝗮𝗯𝗼𝘂𝘁 𝗱𝗲𝗳𝗶𝗻𝗶𝗻𝗴 𝘁𝗵𝗲 𝗿𝗶𝗴𝗵𝘁 𝗴𝗿𝗼𝘂𝗽, 𝗳𝗶𝗻𝗱𝗶𝗻𝗴 𝘁𝗵𝗲 𝗿𝗶𝗴𝗵𝘁 𝗼𝘂𝘁𝗰𝗼𝗺𝗲𝘀, 𝗮𝗻𝗱 𝗳𝗼𝗿𝗺𝗶𝗻𝗴 𝘁𝗵𝗲 𝗿𝗶𝗴𝗵𝘁 𝗿𝗮𝘁𝗶𝗼.
𝗪𝗵𝗲𝗻 𝘆𝗼𝘂 𝗰𝗮𝗻 𝗳𝗶𝗻𝗱 𝘁𝗵𝗲 𝗰𝗼𝗿𝗿𝗲𝗰𝘁 𝗱𝗲𝗻𝗼𝗺𝗶𝗻𝗮𝘁𝗼𝗿, 𝘆𝗼𝘂 𝗵𝗮𝘃𝗲 𝗮𝗹𝗿𝗲𝗮𝗱𝘆 𝘀𝗼𝗹𝘃𝗲𝗱 𝗺𝘂𝗰𝗵 𝗼𝗳 𝘁𝗵𝗲 𝗽𝗿𝗼𝗯𝗹𝗲𝗺.
Other Pages to explore
SAT MATH FORMULA SHEET FOR QUICK REFERENCE
ALGEBRA
SO;VING LINEAR EQUATIONS [PART 1]
SOLVING LINEAR EQUATIONS [PART II]
SYSTEM OF EQUATIONS [ PART I ]
SAT Systems of Equations: Part 2 — Hard Questions, Word Problems, Graphs, Parameters & SAT Tricks
PARAMETER QUESTIONS IN SYSTEMS OF LINEAR EQUATIONS
QUADRATIC EQUATIONS [PART I]
QUADRATIC EQUATIONS [PART II]
PERCENTAGES
PERCENTAGES [introduction]
PERCENTAGE INCREASE AND DECREASE
SUCCESSIVE PERCENTAGES DISCOUNT PROFIT LOSS
