Fractions question! Woo!?

Question

Im sure my revision book has made a typo, the question is, Annie throws a biased dice twice, the probability that she will throw a 6 is 1/4. What is the probability that she she gets at least one 6. My math to solve this was:

(1/4 x 3/4) + (3/4 + 1/4) = 3/16 + 3/16 = 6/16

So my answer was 6/16, but the answer booklet says 7/16, have I gone wrong, or is the answer booklet a little retarded...

Answer 1

The probability she throws at least one six is equal to one minus the probability she throws no sixes.



Probably she throws no six on throw one: 3/4.

Probably she throws no six on throw two: 3/4

Multiply the probabilities to get the probability that she throws no sixes. 3/4 * 3/4 = 9/16.



1 - (9/16) = 7/16.

Answer 2

Your question is confusing. Do the chances 1/4 of throwing a six relate to one throw of the dice or to two? If the latter than the answer is simply 1/4! If the former than the chances of throwing a six must be twice 1/4, i.e. 2/4 or 1/2.

Answer 3

i honestly don't know because when i did it i got 3/8



its simplified.

sorry