Question

Pick a number from a bowl having 1-100 numbers, such that first number, a is always greater than second number,
b. What is the probability of this event

Answer 1

Probability = 1/2 that first number, a is always greater than second number,  if selected from  a bowl having 1-100 numbers

Step-by-step explanation:

Let say Number a selected = 1

Probability of 1  = 1/100

Probability of selecting b < a = 0

Let say Number a selected = 2

Probability of 2= 1/100

Probability of selecting b < a =  1/99    ( as 1 is only smaller than 2 out of remaining 99 numbers)

Let say Number a selected = 3

Probability of 3= 1/100

Probability of selecting b < a =  2/99    ( as 1 & 2 smaller than 3 out of remaining 99 numbers)

and so on

Let say Number a selected = 99

Probability of 99= 1/100

Probability of selecting b < a =  98/99    ( as 98 numbers are smaller than 99)

Let say Number a selected = 100

Probability of 100= 1/100

Probability of selecting b < a =  99/99    ( as all are smaller than 100)

probability of this event that a is always greater

= (1/100) ( 0  + 1/99  + 2/99 +....................................................................+98/99 + 99/99)

= (1/9900) ( 1 + 2 + 3 + ..............................+98 + 99)

= (1/9900) ( 99 *100)/2

= 1/2

Learn More:

The probability for event A is 0.3,

https://brainly.in/question/11925091

In each of 4 different competitions

https://brainly.in/question/12291074