Probability Basics 3

HStatistics.com

Probability Basics 3

Complementary Events:

Having defined an event E, we can also define a Complementary Event - denoted by E' the set of “not desired results" or "unfavourable outcomes".
In the case of rolling a die, for example, if the event E is defined as "getting a composite number",
then E = {4, 6}, while E' = {1, 2, 3, 5}.

Note that all outcomes are either favourable or unfavourable, and hence we can say

n(E) + n(E') = n(S)

If we divide this throughout by n(S), we can say
n(E)/ n(S) + n(E')/ n(S) = n(S)/ n(S)

which gives us
P(E) + P(E') = 1
or,
P(E') = 1 - P(E)

This means that if we have, let's say, 6 possible Mutually Exclusive and Exhaustive outcomes, and we want to find the probability that any one of 5 of these outcomes takes place, we could equally well solve it by finding the probability of the remaining outcome, and subtracting that from 1.

If two dice are rolled, what is the probability that the sum on their faces will be 4 or more?

As we saw last time, the sample space would consist of 36 cases.
Now the sum on the dice could be 4, 5, 6, 7, 8, 9, 10, 11 or 12.
This would mean computing 9 cases and adding them, not a very comforting proposition.
However, we can see that the only possibilities which do not satisfy the given requirement are a sum of 2 or 3.
This gives us only three cases {1, 1}, {l, 2}, {2, 1}
So the probability of the sum being 2 or 3 (i.e. of the required conditions not being met) is 3/36 = 1/12
The required probability would therefore be 1 - 1/12 = 11/12

Lets take another example,

Three friends are planning their birthday parties.
What is the probability that at least one Of them will be on a weekend?

Here, we could have exactly one birthday on a weekend (3 ways, as any one of the three could be the one) or exactly two on weekends (again 3 ways) or all three on weekends (l way) leading to a total of 7 cases to evaluate.
But a simpler exercise would be to find the complementary probability - that none of them will have their birthday on a weekend.
Now each of them has 7 possible days to celebrate their birthday and So the total number of ways for all three will be 7 * 7 * 7 = 343 ways.
Each of them has 5 ways to celebrate their birthday on a non-weekend, so our numerator will be 5 * 5 * 5 = 125.
Hence the probability that none of them will have a birthday on the weekend is 125/343 and the required answer will therefore be 1 - 125/343 = 218/343

Lets go through a few more questions:

Q1. A person writes numbers from 1 to 100 on chits of paper and puts them in a bag. Another person draws two chits of paper one after another without replacement. What is the probability that both numbers are perfect squares?

The probability that the first number is a perfect square will be 10/100 (Since there are 10 perfect squares among the 100 numbers).
Given that the first number picked was a perfect square, there will be only 9 perfect squares left in the remaining 99, and hence the probability that the second number will also be a square will be 9/99.
Therefore, P(1st and 2nd both perfect squares) = P(lst perfect square) * P(2nd perfect square, given 1st perfect square) = 10/100 * 9/99 = 1/110

Q2. A person writes numbers from 1 to 100 on chits of paper and puts them in a bag. Another person draws two chits of paper One after another with replacement. What is the probability that both numbers are perfect squares?

The probability that the first number is a perfect square will be 10/100
Since the chit is being replaced in the bag, the probability that the second number will also be a perfect square will again be 10/100.
Therefore, P(1st and 2nd both perfect squares) = P(lst perfect square) * P(2nd perfect square, given 1st perfect square) = 10/100 * 10/100 = 1/100
In the next post, we shall use the above example to understand the terms Dependent and Independent.