Table of contents | |
What is Probability? | |
Sample Space | |
Sample Space- Examples | |
Some Solved Examples |
A collection of all possible outcomes of an experiment is known as sample space. It is denoted by ‘S’ and represented in curly brackets.
When we toss a coin, there can be only two outcomes i.e., either head or tail. So, the sample space will be, S = {H, T} where H is the head and T is the tail.
When we flip two coins together, we have a total of 4 outcomes. H1 and T1 can be represented as heads and tails of the first coin. H2 and T2 can be represented as heads and tails of the second coin. So, the sample space will be,
S = {(H1, H2), (H1, T2), (T1, H2), (T1, T2)}
With this, we know that if we have ‘n’ coins, the possible number of outcomes will be 2n.
On rolling a die, we can have 6 outcomes. So the sample space will be, S = {1, 2, 3, 4, 5, 6}.
4. Rolling Two Dice Together
When we roll two dice together, we get double the outcomes than when we roll a single outcome. When we roll 2 dice together we get 36 outcomes (6 x 6 = 36).
Note:
- In probability the order in which events occur is important E1 & E3 are treated as different outcomes.
- A pack of playing cards consists of four suits called Hearts, Spades, Diamonds, and Clubs. Each suite consists of 13 cards.
Example 1: A game of chance consists of spinning an arrow that comes to rest, pointing at any one of the numbers such as 1, 2, 3, 4, 5, 6, 7, or 8, and these are equally likely outcomes. What is the probability that it will point at? (i)8 (ii) Number greater than 2 (iii) Odd numbers
Sol:
Sample Space = {1, 2, 3, 4, 5, 6, 7, 8}
Total Numbers = 8
(i) Probability that the arrow will point at 8:
Number of times we can get 8 = 1
P (Getting 8) = 1/8.
(ii) Probability that the arrow will point at a number greater than 2:
Number greater than 2 = 3, 4, 5, 6, 7, 8.
No. of numbers greater than 2 = 6
P (Getting numbers greater than 2) = 6/8 = 3/4.
(iii) Probability that the arrow will point at the odd numbers:
Odd number of outcomes = 1, 3, 5, 7
Number of odd numbers = 4.
P (Getting odd numbers) = 4/8 = ½.
Example 2: A bag contains only lemon-flavoured candies. Arjun takes out one candy without looking into the bag. What is the probability that he takes out an orange-flavoured candy?
Sol:
Let us take the number of candies in the bag to be 100.
Number of orange-flavoured candies = 0 [since the bag contains only lemon-flavoured candies]
Hence, the probability that he takes out an orange-flavoured candy is:
P (Taking orange-flavoured candy) = Number of orange-flavoured candies / Total number of candies.
= 0/100 = 0
Hence, the probability that Arjun takes out an orange-flavoured candy is zero.
This proves that the probability of an impossible event is 0.
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