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# really need help NOW!!! watch

1. i always struggle with probability tree diagrams n never b able to do them
is there any other ways to approach questions like tree diagrams or how to make it easier
2. (Original post by fatima1998)
i always struggle with probability tree diagrams n never b able to do them
is there any other ways to approach questions like tree diagrams or how to make it easier
Practice, do easy questions to warm up then move on to harder ones. Think about what information you're given and what it wants you to find out.

Are there any examples of how you've attempted a question or a question you find difficult?
3. (Original post by SeanFM)
Practice, do easy questions to warm up then move on to harder ones. Think about what information you're given and what it wants you to find out.

Are there any examples of how you've attempted a question or a question you find difficult?
i just get stuck!!! i just dunno how to start them
4. (Original post by fatima1998)
i just get stuck!!! i just dunno how to start them
That one is not so much tree diagrams as immediately identifying the pairs that sum up to 40 or more.
5. (Original post by SeanFM)
That one is not so much tree diagrams as immediately identifying the pairs that sum up to 40 or more.
but it can b attempt as a tree diagram???
6. (Original post by fatima1998)
but it can b attempt as a tree diagram???
I suppose so... With 36(?) branches.
7. (Original post by SeanFM)
I suppose so... With 36(?) branches.
hahaha
then how it can b solved???
i know that tree diagrams always come up in maths exam and i can't do them and lost 4 or 5 marks easily
8. look at how many two coin combinations are given by 10p, 10p, 10p, 20p, 20p, 50p. to find out how many give greater than or equal to 40p, look at the them and see how many pairs you can make with them. e.g. one of the pairs is 20p and 50p, 20p has a probability of 2/6 and 50p has a probability of 1/6. once you find the pairs, work out the probability of getting each one individually. remember you're not putting them back in the box so the amount of coins in the bag changes, hence conditional probability.
9. (Original post by fatima1998)
hahaha
then how it can b solved???
i know that tree diagrams always come up in maths exam and i can't do them and lost 4 or 5 marks easily

See the post above.

If you've got a different question we can work through it together.

Otherwise I'd advise highlighting information in the question, making sure the branches add up to 1, knowing what you're asked to calculate.
10. (Original post by Raees_Sharif)
look at how many two coin combinations are given by 10p, 10p, 10p, 20p, 20p, 50p. to find out how many give greater than or equal to 40p, look at the last three and see how many pairs you can make with them. e.g. one of the pairs is 20p and 50p, 20p has a probability of 2/6 and 50p has a probability of 1/6. once you find the pairs, work out the probability of getting each one individually. remember you're not putting them back in the box so the amount of coins in the bag changes, hence conditional probability.
ok!!! kinda make sense i think!!!
however, i have a question about the three diagram; the video u sent talked about the "with replacement" and "without replacement"
when we should either condition and what are they for??
11. (Original post by SeanFM)

See the post above.

If you've got a different question we can work through it together.

Otherwise I'd advise highlighting information in the question, making sure the branches add up to 1, knowing what you're asked to calculate.
and thanks for ur help too
12. (Original post by fatima1998)
ok!!! kinda make sense i think!!!
however, i have a question about the three diagram; the video u sent talked about the "with replacement" and "without replacement"
when we should either condition and what are they for??
think of it like this; without replacement means you don't put the sweet back in the bag, hence all the probabilities change since you have one less sweet; with replacement means you put the sweet back in the bag, hence the probabilities don't change because you're back with the original quantity before you took the sweet.

the first thing you want to do in a question like this is identify the individual probabilities of each coin before you do anything else. for example, there are 3x 10p coins out of 6 coins so probability of getting a 10p coin is 3/6 or P(10p) = 3/6. once you've done that you need to work out how many combination of coins give you more than or equal to 40p. e.g. 10p + 50p gives more than or equal to 40p so to calculate the probability of getting both, you'd use the without replacement rule; 3/6 x 1/5 = 1/10 (1/6 chance of getting a 50p coin, but 1/5 because we removed a 10p coin from the bag). Or using proper notation it would be P(10p ∩ 50p) = 1/10. Conversely 50p + 10p also gives more than or equal to 40p so you'd do the same again, you don't need to work anything out again since multiplication is commutative 3/6 x 1/5 == 1/5 x 3/6 = 1/10. now do the same for the remaining three combinations and add them all together to get the answer
13. (Original post by Raees_Sharif)
think of it like this; without replacement means you don't put the sweet back in the bag, hence all the probabilities change since you have one less sweet; with replacement means you put the sweet back in the bag, hence the probabilities don't change because you're back with the original quantity before you took the sweet.

the first thing you want to do in a question like this is identify the individual probabilities of each coin before you do anything else. for example, there are 3x 10p coins out of 6 coins so probability of getting a 10p coin is 3/6 or P(10p) = 3/6. once you've done that you need to work out how many combination of coins give you more than or equal to 40p. e.g. 10p + 50p gives more than or equal to 40p so to calculate the probability of getting both, you'd use the without replacement rule; 3/6 x 1/5 = 1/10 (1/6 chance of getting a 50p coin, but 1/5 because we removed a 10p coin from the bag). Or using proper notation it would be P(10p ∩ 50p) = 1/10. Conversely 50p + 10p also gives more than or equal to 40p so you'd do the same again, you don't need to work anything out again since multiplication is commutative 3/6 x 1/5 == 1/5 x 3/6 = 1/10. now do the same for the remaining three combinations and add them all together to get the answer
WOWW
make sense to me
thanks for ur time
14. (Original post by fatima1998)
WOWW
make sense to me
thanks for ur time
did you finish the problem?
15. (Original post by Raees_Sharif)
did you finish the problem?
Yes!!!

n I had my exam too n tbh it was much easier than the one I did in summer GCSES

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