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High School Mathematics - 2
13.3 Sample Space

Sample space

Set of all possible outcomes of an experiment or trial is called sample space and is denoted by S

Consider the following examples

  1. Tossing of a coin
     The outcome will be either heads or tails
          so S = {H , T}
             n(S) = 2 
  2. Tossing of 2 coins
    If 2 coins are tossed each can give two results 
         so  S = { HT , HH , TT, TH}
             n(S) = 4 
  3. Tossing of a dice
         S = { 1,2,3,4,5,6 }
              n(S) = 6 

Events

Consider example 2 stated above
Now we may be interested in some happening. Consider the outcome that the pair has one heads and one tails
so E = { HT , TH ) so n(E) = {2}
Thus the chance that an event happens is called Probability
Formula P(E) = n(E) / n(S)

Solved Examples

  1. 7 books on physics and 4 on mathematics are to be arranged at random on a shelf. Find the probability that the books on maths do not stand next to each other

    Solution

    11 books can be selected in 11P11 ways
    n(S) = 11!
    Let B be the event that no 2 maths book stand next to each other
    Look at the following arrangement
    M P M P M P M P M P M P M P M
    7 books on physics can be arranged in 7P7= 7!ways
    The 4 maths books can take any of the 8 positions shown above.
    n(B) = 7P7x8P4
    P(B) = n(B)/n(S) = 7!x8x7x6x5/11!
    P(B) = 7/33

  2. The letters of the word 'E Q U A T I O N' are arranged at random . Find the probability that the arrangement
    a)Starts with a vowel and ends with a consonant.
    b)Starts with a vowel and ends with a vowel 5/14.

    Solution

    n(s) = 8P8= 8!
    Consider the first case(a)
    Let A be the event that the word starts with a vowel and ends with a consonant.
    As there are 5 vowels in the word the first position can be filled in 5P1=5 ways.
    As there are 3 consonants in the word the last position can be filled in 3P1=3 ways.
    So far we have taken care of 2 positions.
    * - - - - - - *
    Now the 6 spaces inbetween can be filled in 6P6=6! ways.
    So n(A) = 5x6!x3
    So P(A) = 5x6!x3/8! = 15/56.
    Consider the second case (b)
    Let B be the event that the word starts with a vowel and ends with a vowel.
    As there are 5 vowels in the word the first position can be filled in 5P1=5 ways.
    With 4 vowels remaining the last the last position can be filled in sup>4P1=4 ways.
    * - - - - - - *
    Now the 6 spaces inbetween can be filled in 6P6=6! ways.
    So n(B) = 5x6!x4/8!
    So P(B) = 5x6!x4/8!

  3. Tom and Dick play with 2 dice. If Tom throws 9 what is Dick's chance of throwing a higher number?

    Solution

    Total number of outcome of throwing 2 dice = 6x6 = 36. Dick must throw either 10,11,12. This can be done in 5 ways.
    (4+6,5+6,6+4,6+5,6+6)
    Dicks chance of throwing a higher number is 5/36.

  4. A 3 digit number is to be formed by using the digits 1,2,3,4,5,6,7,8,9.(Repetions not allowed). What is the probability that the digit formed is greater than 600.

    Solution

    The possibility of forming 3 digit numbers is 9P3 = 504. n(S)= 504. Let A be the event that the number is greater 600.
    n(A)= 4x8P2 = 224.
    P(A)= 224/504.

  5. A bag contains 3 white , 4 black and 2 red balls. 2 balls are chosen at random. What is the probability that 1 white and 1 red ball is chosen.

    Solution

    2 balls can be drawn from 9 balls in 9C2 ways = 36 ways n(S)= 36 n(A)= 3C1.2C1 = 6 P(A) = 6/36 = 1/6


Q 1: A card is drawn from a well shuffled pack of 52 cards. The probability that the the card is a diamond is
1/13
1/3
1/8
1/4

Q 2: If E is an event of sample space S then
P(E) < 0
P(E) > 1
1 < P(E) < 2
0 < P(E) < 1

Q 3: Consider the tossing of a dice . If B is the event that the number is divisible by 2 as well as 3 then n(B) is
3
6
2
1

Q 4: Consider the sample space S= ( 1 ,2,3,4,5,6,7} . Find the event E such that the numbers are all odd numbers . Also find n(E)
E={1,7} n(E) = 4
E= {3 ,4, 5, 6} n(E) = 4
E={1,3,5,7} n(E) = 4
E={3,5,7} n(E) = 3

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