Problem Solving And Data Analysis: Probability

PROBABILITY:

Probability is sometimes referred to as “Chance” in questions on the SAT. It is a measure of the likelihood of a particular outcome when there are several possible outcomes of an event.

Example 1:  The outcome of writing a test is getting a score. If we consider the math section of the SAT, the score could be any of the multiples of 10 from 200 to 800 (61 possible different scores). Now, the probability of getting a particular score, say 740 would be 1 out of 61 or 1/61.

Formula 1:

Probability = No. of ways a favoured outcome can result / Total possible outcomes.

Example 2:  What are the chances that a randomly chosen year is a leap year?

Solution:  Going by the formula given above, the chances/probability of a randomly chosen year being a leap year can be obtained as follows:

The denominator in the formula (total possible outcomes) here is a little tricky one to handle, as the size of the sample (total number of years) is not specified.

However, if we work with a particular sample size, let’s say, the years from 0001 to 2016, then there are 2016 possible years that one can choose from. Hence, the denominator in this case would be 2016.

Now, the numerator is easy to understand in this case as every fourth year (0004, 0008,……2012,2016) is a leap year. So we have 2016/4 = 504 leap years from year 0001 to the year 2016. Hence the numerator would be 504.

The probability in this case as per the formula would be 504/2016 = 1/4 or 0.25

Important Note: If the number of years available to be chosen from was given in the question, the answer would be different as a time period was specified.

Similarly, the probability of any outcome is calculated by dividing the number of ways in which the outcome can result by the number of total possible outcomes.

Example 3: 3 six sided dice numbered 1 to 6 on their faces are rolled. What are the chances that all the three show up an odd number on the top?

Solution: Applying the formula mentioned above, the number of ways in which the favoured (odd on all the three dice) outcome results = 27

How do we get this figure: (The first dice can show up either a 1 or a 3 or a 5, hence 3 favourable outcomes, similarly the second and the third dice can result in 3 favourable outcomes each. Now, the total favourable outcomes would be the product of the individual favourable outcomes = 3 * 3 * 3 = 27

Similarly, the total number of outcomes = 6 * 6 * 6 = 216.

Now, the required probability would be = 27/216 = 1/8 or 0.125

Important Note: Observe that the probability value can be expressed as a fraction or a decimal whose value is always greater than or equal to 0 and less than or equal to 1.

Most of the probability questions on the SAT math test you on the application of the above formula. However, there are a few more formulae you are supposed to know:

Probability of ‘not resulting of an outcome’ equals Probability of that outcome subtracted from 1.

Formula 2:

P (Not A) = 1 – P (A)

Let’s understand the rationale behind the above formula.

Example 4:  3 six sided dice numbered 1 to 6 on their faces are rolled. What are the chances that not all the three dice show up an odd number on the top?

Solution: Now, if we wanted to calculate the probability of not all the three dice showing up odd numbers on the top, we need to carefully understand the meaning of the word  “not all”.

“Not all” means not all the three are odd, however, 1 of them or 2 of them can be odd. Hence, the required probability here is of all the three dice together not showing up odd numbers on the top.

Let’s work with the favourable outcomes:

The outcomes in which one of them is odd and the other two are even = 3 * (3*3*3)

(3*3*3 for the first dice to show up an odd number (1 or 3 or 5) and the remaining two showing up an even number (2 or 4 or 6)), (3 times all of that because any of the three dice could be the one showing up the odd number)

Similarly, the outcomes in which one of the is even and the other two are odd = 3 * (3*3*3)

Also, the outcomes in which all the three are showing even on the top are also favourable = 3*3*3.

Hence, the total number of favourable outcomes is the sum of all the above = (3*3*3*3) + (3*3*3*3) + (3*3*3) = 81+81+27 = 189.

The total number of possible outcomes remains the same = 6*6*6 = 216

Hence the required probability would be = 189/216 = 7/8 = 0.875

This is also got applying the second formula introduced:

Probability of (Not A) = 1 – Probability (A).

Here, in the above example, it would be:

P (Not all Odd) = 1 – P (All odd) = 1 - 1/8 = 7/8.

Let us look at two more formulae….

Formula 3: Probability of A & B is the product of the individual probabilities of A and B.

P (A&B) = P (A) * P (B)

Formula 4: Probability of A or B, where A and B refer to a favourable outcome equals the sum of the individual probabilities of A, B minus the Probability of both A&B as an outcome.

P(A or B) = P(A) + P(B) – P(A&B)

Now let us understand the above 2 formulae using the following examples.

Example 5: When 2 six sided dice are rolled, what are the chances that they show up an odd number on one of them and an even number on the other.

Solution: Using the formula P (A & B) = P (A) * P (B),

Probability          = [P (odd on the first) * P (Even on the second)] + [P (even on the first) * P (Odd on the second)]

= (1/2 * 1/2) + (1/2 * 1/2)

= 1/4 + 1/4

= 1/2 = 0.5

Applying the first formula: The favourable outcomes are: (1 on the first along with 2 or 4 or 6 on the second, 3 on the first along with 2 or 4 or 6 on the second, 5 on the first along with 2 or 4 or 6 on the second, 2 on the first with 1 or 3 or 5 on the second, 4 on the first with 1 or 3 or 5 on the second, 6 on the first with 1 or 3 or 5 on the second)

Hence, the number of favourable outcomes equals 3+3+3+3+3+3 = 18

And the number of total possible outcomes equals = 6*6 = 36.

Required Probability = 18/36 = 0.5

Example 6: What is the probability of a 6 sided dice showing up an even number or multiple of 4 on the top?

Solution: Using the formula P (A or B) = P (A) + P (B) – P (A&B),

Required probability= P (even) + P (multiple of 4) – P (even & multiple of 4)

= 3/6 + 1/6 –1/6 = 3/6 = 0.5

Applying the first formula: The favourable outcomes are: 2 on the top, 4 on the top and 6 on the top. These are three in number.

The total outcomes are 6. Hence, the required probability is again 3/6 = 0.5

Summary:

To summarize so far, most of the questions on the topic of probability test you on the following three formulae:

1. P (A) = Number of ways a favourable outcome can result / Total number of possible outcomes
2. P (A or B) = P (A) + P (B) – P (A & B)
3. P (A & B) = P (A) * P (B)

Also, note that the probability values are always expressed either as fractions, decimals or percentages, all of which are always greater than or equal to 0 and less than or equal to 1 in value.

Example: 1/2 or 0.5 or 50%, 1/4 or 0.25 or 25%, 3/10 or 0.3 or 30% and so on.

You may now try to solve:

If the probability of Fritz completing a task is 1/7, the probability of Gerald completing the same task is 3/7, then what is the probability that either Fritz alone or Gerald alone to complete the task?

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