Test: Axiomatic Probability - Commerce MCQ

Equally Likely Events Events which have the same chance of occurring Probability. Chance that an event will occur. Theoretically for equally likely events, it is the number of ways an event can occur divided by number of outcomes in the sample space.

The cards contains 4 Queen from which 2 are black and 2 are red
we need to find the probability that the card drawn is red and a queen is: 2/52
= 1/26

The probability on the basis of observations and collected data is called statistical approach of probability.

Sample space = {HH, HT, TH, TT}
n(SS) = 4 
No tail = {HH}
n(No tail) = 1 
P(No tail) = n(No tail) / n(SS
= 1/4   

Possible outcomes :
{1,1};{1,2};{1,3};{1,4};{1,5};{1,6}
{2,1};{2,2};{2,3};{2,4};{2,5};{2,6}
{3,1};{3,2};{3,3};{3,4};{3,5};{3,6}
{4,1};{4,2};{4,3};{4,4};{4,5};{4,6}
{5,1};{5,2};{5,3};{5,4};{5,5};{5,6}
{6,1};{6,2};{6,3};{6,4};{6,5};{6,6}
Total outcomes = 36
Favorable events : an even number as the sum :{1,1};{1,3};{1,5};{2,2};{2,4};{2,6};{3,1};{3,3};{3,5};{4,2};{4,4};{4,6};{5,1};{5,3}{5,5};{6,2};{6,4};{6,6} =18
 
So, the probability of getting an even number as the sum:
= Favorable events/Total events
= 18/36
 = 1/2

Since 1 ticket is choosen out of 2000 tickets
n(S) = ²⁰⁰⁰C₁
= (2000!/1!1999!)
= 2000
Now out of 2000 tickets only 50 have a prize
Hence no of tickets not having prize
= 2000 - 50
= 1950
Let A be the event that if we buy 1 ticket it doesnt have a prize
Hence, 1 ticket will be out of 1950 tickets
n(A) = ¹⁹⁵⁰C₁
= 1950
Probability not getting a prize if we get one ticket P(A) = n(A)/n(S) 
= 1950/2000

When a coin is tossed. Head or Tail may occur. Whereby when a die is thrown the numbers from 1 & 6 may occur.
∴ The sample space S={H₁T₁, H₂T₂, H₃T₃, H₄T₄, H₅T₅, H₆T₆}
H represents Head.,  
H₁ represents head in coin and 1 in dice
T represents Tail.