Probability Basics 2

# HStatistics.com

### Probability Basics 2

The third condition we established (in the previous post) for a sample space (equiprobability of the elements) can sometimes lead to slightly counter-intuitive results.
We need to understand the underlying logic carefully. So let's spend a bit of time examining it more closely.

For example, an unwary person, when raced with a question like "what is the probability that Virat will score a century in the next match?" will say "Well, either he will score a century, or he won't, and therefore the probability should be 1/2 right?"

However, one could refute that in several ways.

For example:
"What would be the probability that Virat will score 500 in his next match?"
OR
"But you could also say that he could score O, or 1, or 2, or. ...or 100. So that gives us 101 cases. So should it be 1/101? And in fact, he could go on to score more than 100 as well..."
OR
"So then, what is the probability of Ishant scoring a century in his next match (not in bowling, mind!)?"
The problem being of course, that scoring a century does not have the same likelihood as that of not doing so.
In fact, even the great Don Bradman scored a century every 3 innings or so and so in any given innings his probability of scoring a century would have been a bit above 1/3.
In fact, in real life conditions, this is how probability is usually estimated - using historical data. And hence, while some questions will directly tell you the probability, some will be presented in the form of data.
So if a question has a sentence like "2 out of every 5 bulbs made by company X are defective"
or "Robin Hood hits the target three times out of every four arrows he shoots" we should understand that this does not mean that in the next 5 bulbs you pick, 2 will be faulty or that in the next four arrows shot by Robin, three will hit the target. Rather,
we should understand that for each bulb, the probability that it is faulty will be 2/5 and similarly for each arrow shot, the probability that it hits is 3/4.

Consider two coins being tossed. What the probability Of getting two heads?

There could be two ways of generating a Sample Space:
b){H,H}, {H,T}, {T,H}, {T,T}

Now the first would give us a probability of 1/3 while the second would give us a probability of 1/4.
Both cannot simultaneously be correct!
But on closer examination we find that {Two Heads} can occur in only one way while
{One Head and One Tail} can occur in two ways and hence they are not equiprobable - this invalidates the first sample space.
The correct answer will be 1/4 and this will be true whether the coins are distinct or identical.

Let's take a few examples now.

Example 1:
If three identical coins are tossed, what is the probability of getting 2 Heads?

The valid Sample Space would consist of the following 8 possibilities:
{H, H, H} ,{H, H, T}, {H, T, H} , {T, H, H}, {H, T, T}, {T, H, T}, {T, T, H}, {T, T, T}
Case number 2,3 and 4 would be favourable.
Hence the required probability will be 3/8

Example 2:
If two dice are rolled, what is the probability that the sum on their faces will be 8?

(Note that an erroneous sample space here would be {Sum = 2}, { Sum = 3}.. {Sum = 12}
The valid sample space will consist of 36 cases. {(1,1), (1,2)...so on till (6,6)}
There will be 6 ways for each die and hence a total of 6 x 6 = 36 cases (applying the Fundamental Principle of Counting) you should not need actual listing!):

The favourable cases would be (2,6),(3,5),(4,4),(5,3),(6,2).
Thus, we can observe that the required probability will be 5/36.

Example 3:
If three dice are rolled, what is the probability that the sum on their faces will be 16?

We can easily work out that the sample space will consist of 6 x 6 x 6 — 216 cases.
We can also carefully list out the favourable cases (which will add up to 16) as follows:
{6, 6, 4}, {6, 5, 5}, {5, 6, 5}, {5, 5, 6}, {4, 6, 6}
Hence the required probability will be 6/216 = 1/36.

Example 4:
If two dice are rolled, what is the probability that the sum on their faces will be 10 or more?

As above the sample space would consist of 36 cases.
Now the sum on the dice could be 10 {6, 4}, {5, 5}, {4, 6}, or 11 {5, 6}, {6, 5}, or 12 {6, 6}.
So the required probability would be 6/36 or 1/6.