Question

There are 100 prize tickets in a bowl, numbered 1-100. What is the probability that an even numbered prize will be chosen at random, not replaced, then an odd numbered prize ticket will be chosen?

Answer 1

Probability that it is an even number:

• Total outcome = 100
• Favourable outcome = 50 (because there are 50 even numvers and 50 odd numbers)

`\tt \: P(E) = (F.O.)/(T.O.)`

`\tt \: P(E) = (50)/(100) = (1)/(2)`

Pribability that it is an odd number:

• Total outcome = 99 (because one ticket was taken out and was not replaced)
• Favourable outcome = 50

`\tt \: P(E) = (F.O.)/(T.O.)`

`\tt \: P(E) = (50)/(99) = 0.505051`

Answer 2

25/99

Explanation:

If the tickets are numbered 1-100, half of the tickets will be even and half of the tickets will be odd

Number of even tickets = 50

Number of odd tickets = 50

Let A be the event => even ticket on first draw

Let B be the event => odd ticket on second draw

P(even on first draw) = P(A)
= Number of even tickets/total number of tickets

= 50/100

= 1/2

Once a ticket has been drawn and the second draw is without replacement,

Total number of tickets remaining = 100 - 1 = 99

Total number of odd tickets remaining given first ticket drawn is even
= 50

P(odd ticket second draw | first draw is even) = P(B | A)
= 50/99

P(A and B) = P(A) · P(B | A)

= 1/2 x 50/99

= 50 / (2 x 99)

= 50/198

= 25/99