For a particular species of cat, the odds against a kitten

For a particular species of cat, the odds against a kitten being born with either blue eyes or white spots is 3:1. If the probability of a kitten exhibiting only one of these traits is equal and the probability of exhibiting both the traits it 10%, what are the odds in favour of a kitten having blue eyes?

The answer is given as being 7:33, but how would I arrive at this answer?

Thank you.

Re: PROBABILITY

cr7 said:

for a particular species of cat, the odds against a kitten being born with either blue eyes or white spots is 3:1. if the probability of a kitten exhibiting only one of these traits is equal and the probability of exhibiting both the traits it 10%, what are the odds in favour of a kitten having blue eyes?
the answer is 7:33
how do u get this answer?

Click to expand...

change your odds into a probability and use


P(B or S)=P(B)+P(S)-P(B and S)

thanx ... can u actually show a full solution because i tried that and i dun get 7:33 please and thanks

cr7 said:

thanx ... can u actually show a full solution because i tried that and i dun get 7:33 please and thanks

Click to expand...

How about you show us your work, so we don't change you method, thus causing less confusion? :idea:

1/4= 2P(B)- 1/10
1/4+1/10=2P(B)
7/20=2P(B)
P(B)=7/10

i am stuck int he same question i get 1/4 but the plugging in of numebrs into the formula is what i don't get, why is 1/4 = P(B or S)?? i understand 10% = P(B and S) and also how do you arrive to 2P(B)???

Considering to use the non-mutually exclusive event formula
P(B or W) = P(B) + P(W) - P(B and W)
P(B or W) =1/4 because of the odds in favour of their occurrence prob are 1/4( odds against are 3:1, then odds in favour would be 1:3. and turn it to prob form, 1/4)
Since we know their chance is the P(B) and P(W) is the same they are both 1/2,
so P(B) + P(W) =2P(B) or 2P(W), and knowing that we are only looking for Blue eyes, so we see it as 2P(B)
The information also mentions that the probability for them to exhibiting both traits are 10%(10/100 = 1/10), fits to (P and B),
now bring everything on the board
1/4 = 2P(B) - 1/10
1/4 + 1/10 = 2P(B)
1/2(1/4+1/10)= P(B)
1/2(7/20) = P(B)
7/40 = P(B)
turn it to odds form,
odds in favour of blue eyes = 7:33

Thanks for the answer. Do you realize that this post was from 2006?