A number from 1 to 10 is chosen at random. What is the probability of choosing a 5 or an even number?
The numebers chosen are from 1 to 10.
Then possible outcomes are: 1,2,3,4,5,6,7,8,9,10
The probability of chosen a 5 OR and even number = probability of chosen a 5 + probability of chosen an even number.
Let us calculate each event:
P ( 5) = 1/10
Even number are: 2,4,6,8,10
==> P(even) = 5/10
P( 5 OR even) = P(5) + P(even)
= 1/10 + 5/10 = 6/10
==> P ( 5 or even) = 3/5
We can chose randomly a number from 1 to 10 in 10 ways, as any of the number in a random choice is equally likely.
Number of ways of getting particular number 5 in a random choice of numbers from 1 to 10 is only one way.
The number of ways of getting an even number (evens are 2,4, 6,8 and 10) = 5 ways.
Therefore the probablity of getting a 5 or an even number in a choice = number ways of getiing 5 or an even number / total number of ways of chosing any number = (1+5)/10 = 0.6.