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1. Dear friends,

I'm unable to solve the following probability question. Please help me solve it. Thanks in advance. The answer given in the book is: 5/9 [for part (b)]. Don't know even if the answer is correct.

Suzi has taken up golf, and she buys a golf bag containing five different clubs. Unfortunately she does not know when to use each club, and so chooses them randomly for each shot. The probabilities for each shot that Suzi makes are shown below

Right club
--------------
Good shot - 2/3

Wrong club
-----------------
Good shot - 1/4

a) Use the above information to construct a tree diagram.
b) At one short hole, she can reach the green in one shot if it is 'good'. If her first shot is 'bad', it takes one more 'good' shot to reach the green. Find the probability that she reaches the green in at most two shots.
2. (Original post by tantrik)
Dear friends,

I'm unable to solve the following probability question. Please help me solve it. Thanks in advance. The answer given in the book is: 5/9 [for part (b)]. Don't know even if the answer is correct.

Suzi has taken up golf, and she buys a golf bag containing five different clubs. Unfortunately she does not know when to use each club, and so chooses them randomly for each shot. The probabilities for each shot that Suzi makes are shown below

Right club
--------------
Good shot - 2/3

Wrong club
-----------------
Good shot - 1/4

a) Use the above information to construct a tree diagram.
b) At one short hole, she can reach the green in one shot if it is 'good'. If her first shot is 'bad', it takes one more 'good' shot to reach the green. Find the probability that she reaches the green in at most two shots.
What have you tried??
3. To set this up; the probability she picks a good club is 1/5 and the probability of a bad club is 4/5.
See if you can finish off the diagram from there. Is this S1?

btw I have solved the question and the answer is correct
4. I could draw the tree diagram which I attached with this reply. Not sure whether the tree diagram is correct. Still I don't know which outcomes should I take for finding the solution to part (b). Let me know what to do next. The question is from Edexcel IGCSE Maths B student book.

Attached Files
5. Tree diagram for part (a).doc (48.5 KB, 35 views)
6. (Original post by tantrik)
I could draw the tree diagram which I attached with this reply. Not sure whether the tree diagram is correct. Still I don't know which outcomes should I take for finding the solution to part (b). Let me know what to do next. The question is from Edexcel IGCSE Maths B student book.

You only need the first two columns of that tree because the last columns implies that good/bad shots are dependent on the previous shots being good/bad which is not the case. Good/bad shots depends on the RIGHT/WRONG club.

1 - So your diagram is correct, ignoring the last column. Now collect the probabilities of getting a GOOD shot because she will reach green after a single GOOD shot.

2 - Then you are told that "if the first shot is bad, then it takes one more GOOD shot to reach the green" therefore reaching green is dependent on getting a GOOD shot AFTER a BAD shot. So multiply the probability of getting a first BAD shot overall with the probability of getting a GOOD shot the second time round.

Sum probabilities from 1 and 2 then there you have it.
7. (Original post by RDKGames)
You only need the first two columns of that tree because the last columns implies that good/bad shots are dependent on the previous shots being good/bad which is not the case. Good/bad shots depends on the RIGHT/WRONG club.

1 - So your diagram is correct, ignoring the last column. Now collect the probabilities of getting a GOOD shot because she will reach green after a single GOOD shot.

2 - Then you are told that "if the first shot is bad, then it takes one more GOOD shot to reach the green" therefore reaching green is dependent on getting a GOOD shot AFTER a BAD shot. So multiply the probability of getting a first BAD shot overall with the probability of getting a GOOD shot the second time round.

Sum probabilities from 1 and 2 then there you have it.
Thanks for your reply. I'm confused on what would be the values for P(right club) and P(wrong club). That is where I'm stuck at now.
8. (Original post by tantrik)
Thanks for your reply. I'm confused on what would be the values for P(right club) and P(wrong club). That is where I'm stuck at now.

As for any given shot you pick either the right one or the wrong one.
9. (Original post by RDKGames)

As for any given shot you pick either the right one or the wrong one.
I tried using your suggestion but I ended up getting a different answer. I tried the following:

P(at most two shots) = (1/5)×(2/3)+(4/5)×(1/4)+(1/5)×(1/3)×(2/3)+(4/5)×(3/4)×(1/4) = 19/36

Where have I made the mistake? Let me know the mistake that I made.

10. (Original post by tantrik)
I tried using your suggestion but I ended up getting a different answer. I tried the following:

P(at most two shots) = (1/5)×(2/3)+(4/5)×(1/4)+(1/5)×(1/3)×(2/3)+(4/5)×(3/4)×(1/4) = 19/36

Where have I made the mistake? Let me know the mistake that I made.

I agree with the underlined portion because this is the probability of getting it in 1 shot.

The second part is trickier and the other half of your calculation falls on it. You need to get the overall probability of getting a BAD shot, then multiply it by the overall probability of getting a GOOD shot as this is now conditional.

If you still do not get this, please show me your probabilities for each of these clearly.
11. (Original post by RDKGames)
I agree with the underlined portion because this is the probability of getting it in 1 shot.

The second part is trickier and the other half of your calculation falls on it. You need to get the overall probability of getting a BAD shot, then multiply it by the overall probability of getting a GOOD shot as this is now conditional.

If you still do not get this, please show me your probabilities for each of these clearly.
I could get the right answer this time. Still I want to know why should I take overall probabilities of BAD shots and GOOD shots (add probabilities of BAD shots and add probabilities of GOOD shots) and then multiply. A little clarification would help.

Thanks
12. (Original post by tantrik)
I could get the right answer this time. Still I want to know why should I take overall probabilities of BAD shots and GOOD shots (add probabilities of BAD shots and add probabilities of GOOD shots) and then multiply. A little clarification would help.

Thanks
Ah good.

The first part simply deals with getting a GOOD shot the first time around. So you collect the overall probability of getting a good shot and that is straight forward.

With the second part, we need to multiply the overall probabilities of BAD shots following by GOOD shots because we are given a specific condition therefore this becomes an extension of our initial conditional probability. As the two events of GOOD/BAD shots are independent of each other, we know that which means that to get the probability of a bad AND THEN good shot (at most 2, as required by the question) we simply multiply their overall probabilities together. This has got to be strictly bad shots followed by good shots to make the 2 shots overall, if there were 2 bad shots and 1 good shot then that is 3 shots overall which is outside our wanted range.

At the end we add the two probabilities from first part and second part to get the grand probability of getting green at most 2 shots. Get it?
13. (Original post by RDKGames)
Ah good.

The first part simply deals with getting a GOOD shot the first time around. So you collect the overall probability of getting a good shot and that is straight forward.

With the second part, we need to multiply the overall probabilities of BAD shots following by GOOD shots because we are given a specific condition therefore this becomes an extension of our initial conditional probability. As the two events of GOOD/BAD shots are independent of each other, we know that which means that to get the probability of a bad AND THEN good shot (at most 2, as required by the question) we simply multiply their overall probabilities together. This has got to be strictly bad shots followed by good shots to make the 2 shots overall, if there were 2 bad shots and 1 good shot then that is 3 shots overall which is outside our wanted range.

At the end we add the two probabilities from first part and second part to get the grand probability of getting green at most 2 shots. Get it?
Yes, thanks for the explanation and clarification.

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