" In a donkey Derby event, there are three races. There are six donkeys entered for the frist race, four for the second and three the third. Sheils places a bet on one donkey in each race. She knows nothing about donkeys and chooses each donkey at random,
Find the probablity that she backs at least one winner."
My initial method, was to draw a tree diagram, to consider the outcomes. But I do not believe my tree diagram makes enough sense. I merely drew one just to consider what the outcomes could be.
I drew 6 branches for the first race, and supposing one donkey came first in first race, I drew 4 branches coming off the first branch considered in the first race. These 4 branches would repesenret outcomes for the second race. Then cosidered one of the 4 donkeys in those four brounces won, I would draw three more branches on that branch for the third race.
P (A donkey wins first race) = 1/6
P (A donkey wins second) = 1/4
P(A donkey wins third race) = 1/3
From the information quoted, I understood For a donkey to win overall, the conditions are : the donkey cannot come 6th or 5th in the first race, the donkey cannot come 4th in the second race, and the Donkey must come 1st in the third race.
I then realised (correct me If Im wrong) , for the First and Second Race it doesnt matter who wins, Rather a probalitliy should be considered of the chances of passing for the first, and second race, if Im correct. So the P( of winning first and second races) are not as useful as considereing the probability of passing the first and second round.
Winner donkey:
Could come 1st 2nd 3rd 4th First race
1st 2nd 3rd Second Race
Must be 1st Third Race.
But I am unsure how to display this on a tree diagram, and thus calculate the probabilities as such.
Find the probablity that she backs at least one winner."
My initial method, was to draw a tree diagram, to consider the outcomes. But I do not believe my tree diagram makes enough sense. I merely drew one just to consider what the outcomes could be.
I drew 6 branches for the first race, and supposing one donkey came first in first race, I drew 4 branches coming off the first branch considered in the first race. These 4 branches would repesenret outcomes for the second race. Then cosidered one of the 4 donkeys in those four brounces won, I would draw three more branches on that branch for the third race.
P (A donkey wins first race) = 1/6
P (A donkey wins second) = 1/4
P(A donkey wins third race) = 1/3
From the information quoted, I understood For a donkey to win overall, the conditions are : the donkey cannot come 6th or 5th in the first race, the donkey cannot come 4th in the second race, and the Donkey must come 1st in the third race.
I then realised (correct me If Im wrong) , for the First and Second Race it doesnt matter who wins, Rather a probalitliy should be considered of the chances of passing for the first, and second race, if Im correct. So the P( of winning first and second races) are not as useful as considereing the probability of passing the first and second round.
Winner donkey:
Could come 1st 2nd 3rd 4th First race
1st 2nd 3rd Second Race
Must be 1st Third Race.
But I am unsure how to display this on a tree diagram, and thus calculate the probabilities as such.
