Not all probability questions on the SAT will be about dice, playing cards and marbles. Here are examples of other possible scenarios.
Example
A study found that 80% of drivers, who ride with a passenger, wear seatbelts. Passengers wear seatbelts 70% of the time when the driver wears a seatbelt, and 55% of the time when the driver doesn’t wear a seatbelt. What is the probability that a passenger wears a seatbelt?
Solution
You can use a tree diagram to visualize the possibilities. Write the probabilities on each branch.
There are 2 paths that give the desired outcome of a passenger wearing a seatbelt. Find and add the probabilities of each path.
(Path 1) Driver wears and passenger wears: 80% × 70% = 56%
(Path 2) Driver does not wear and passenger wears: 20% × 55% = 11%
The probability is 56% + 11% = 67%.
A passenger wears a seatbelt 67% or about 2/3 of the time.
Example
In a 1-mile race, three different schools (Washington High, Duke High and Cherry Hill High) each have 5 competitors. What is the probability that a student from Cherry Hill will take first place, a student from Duke will take second place and another student from Duke will take third place?
Solution
This question has both “or” and “and” in the same problem: independent and dependent.
Since 2 students from Duke are desired outcomes, those are dependent events.
Find each probability.
Cherry Hill student first: 5/15 = 1/3
Duke student takes second: 5/14
Duke student takes third: 4/13
Since this is an “and” question, multiply the probabilities. Don’t forget to factor and reduce before you multiply.
\dfrac{1}{3} × \dfrac{5}{14} × \dfrac{4}{13} = \dfrac{5 × 2 × 2}{3 × 2 × 7 × 13} = \dfrac{10}{273}Example
At 3 p.m. Jennifer went into labor. There is a 0.7 chance her baby will be born during each hour that she is in labor. What is the probability that her baby was born at 5:30 p.m. on the same day?
A) 0.027
B) 0.063
C) 0.147
D) 0.27
E) 0.343
Solution
This question is an independent probability question in disguise. The probability that her baby will be born each hour does not change. Each hour, there is a 0.7 chance the baby will be born, which means there is a 0.3 chance the baby will not be born.
Method 1
From 3 p.m. to 4 p.m., the baby was not born and the probability of that happening is 0.3.
From 4 p.m. to 5 p.m. the probability also is 0.3.
From 5 p.m. to 6 p.m., when the baby was born, the probability is 0.7.
Since the baby must not be born in the first or second hour, and must be born in the third hour, this is an “and” question, so the probability is (0.3)(0.3)(0.7) = 0.063
The correct answer is choice (B).
Method 2
You can use a tree diagram to visualize the possibilities. Write the probabilities on each branch.
There is just one path that gives the desired outcome.
(0.3)(0.3)(0.7) = 0.063
The correct answer is choice (B).
Example
From a class of 12 students, two students will be randomly chosen simultaneously. If g is the number of girls in the class, what is the value of g?
(1) The probability that two girls will be chosen together is 1/11.
(2) The probability that one boy and one girl will be chosen is 16/33.A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
Solution
To answer this question, you don’t need to do the math.
Look at Statement (1). You know there is a specific number of girls (g). Since each number of girls yields a different probability of choosing 2 girls, there must be only one specific number that would yield 1/11. So this is enough information to know the value of g, the number of girls.
Statement (2) requires a little more thought. Pairing a boy and a girl can be (b, g) or (g, b). As in many pairings, there are 2 different pairs that will give the same probability. Statement (2) is not sufficient.
(Note: If you make the table, you will see that there are two ways to get 16/33, with 4 boys and 8 girls or with 4 girls and 8 boys.)
Since only Statement (1) gives enough information, the correct answer is choice (A).
Example
In a hotel with single rooms and double rooms, what is the probability that a room chosen at random will be a double room painted red?
(1) 1/6 of the rooms in the hotel are painted red.
(2) 2/3 of the hotel’s rooms are double rooms.A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
Solution
Statement (1) gives the fraction of rooms painted red, but does not say anything about double rooms.
Statement (2) gives the fraction of the rooms that are double rooms, but does not say anything about red rooms.
Putting the statements together still does not give enough information. Neither statement says how many of the red rooms are double rooms. For example, if there were 12 rooms you would know that 2 were red and 8 were doubles. But you don’t know if any of the doubles are red. There is no information connecting the two categories, so you cannot find the probability.
The correct answer is choice (E).
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