Mathematics 4024 question on probabilities

[ailg1996]

Q1: The ticket machine in a carpark takes 50c coins and $1 coins.Aticket costs 1.5 $.The probability that a machine will accept a particular 50 cent coin is 0.9 and that will accept a 1$ coin is 0.8.
i: Leslie put one 50cent coin and one $1 coin in the machine.Calculate the probability tht the machine will not accept either of these coins.

 

[abcde]

AOA!
P ( machine will not accept a 50 cent coin) = 1 - 0.9 = 0.1
P (machine will not accept a $1 coin) = 1 - 0.8 = 0.2
=> P (machine will not accept either of these coins) = 0.1 x 0.2 = 0.02.
I hope this is correct and that you understand.

 

[Silent Hunter]

wont it be 0.1+0.2? - it wont accept either $1 or 50 cents? :?

 

[leosco1995]

No, abcde is right. The question says 'Calculate the probability tht the machine will not accept either of these coins.' which means that the machine shouldn't accept the 50 cent coin and the $1 coin. If you replace and with or, then I am pretty sure you will add them instead.

 

[abcde]

Agreed with leosco1995.

 

[Silent Hunter]

got it .........thanks