Question

Ann wants to buy a coffee machine. She knows, that during the holidays, the probability of a discount of more than 10% is 0.87, on weekends, the probability of a discount of more than 10% is 0.38, and on a regular weekday, it is 0.17. On average there are 15 holidays per year, 100 weekends (holiday weekends are excluded) and 250 weekdays. Find the probability of a discount of more than 10% for the randomly chosen day. Find the probability that it was a weekend if it is given that she bought a machine with a discount of more than 10%.

Answer 1

1. 0.256

2. 0.406

Explanation:

Total days:

15 + 100 + 250 = 365

P(more than 10% discount)

= (0.87×15/365) + (0.38×100/365) +(0.18×250/365)

= 1871/7300

0.2563013699

P(weekend/more than 10% discount)

= (0.38×100/365) ÷ (1871/7300)

= 760/1871

0.4061998931

Answer 2

Let's denote:

P(discount on 15 holidays) = P(A) = 0.87

P(discount on 100 weekends) = P(B) = 0.38

P(discount on 250 weekdays) = P(C) = 0.17

=> The probability of a discount of more than 10% for the randomly chosen day: P = P(A) x 15/360 + P(B) x 100/360 + P(C) x 250/360

= 0.87 x 15/360 + 0.38 x 100/360 + 0.17 x 250/360

= 0.26

Given she bought a machine with a discount of more than 10%, the probability that it was a weekend would be:

P = 100/(100 + 250 + 15) = 0.27

Hope this helps!

:)