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# S1 question watch

1. Jenny and Omar are each allowed two attempts at a high jump.

(i) The probability that Jenny will succeed on her first attempt is 0.6. If she fails on her first attempt, the probability that she will succeed on her second attempt is 0.7.
Calculate the probability that Jenny will succeed.

(ii) The probability that Omar will succeed on his first attempt is p. If he fails on his first attempt, the probability that he will succeed on his second attempt is also p. The probability that he succeeds is 0.51. Find p.
2. Is this a probability distribution?
3. (i) 0.6+((1-0.6)*0.7) Succeed+ Fail then Succeed
(ii)x+((1-x)*x)=0.51 so 2x-(x^2)=0.51 Then solve as quadratic.
4. (Original post by jameswhughes)
(i) 0.6+((1-0.6)*0.7) Succeed+ Fail then Succeed
(ii)x+((1-x)*x)=0.51 so 2x-(x^2)=0.51 Then solve as quadratic.
I missed a lesson on this. Why is it 1 - p exactly?

This is a Geometric distribution right?
5. (Original post by No!)
I missed a lesson on this. Why is it 1 - p exactly?

This is a Geometric distribution right?
Yes, geometric distribution. The probability of failing is the probability of not succeeding, hence the 1-p.
6. I got P = 0.88

and 0.51 = P^2

Not sure about either tbh, I haven't revised this part of stats yet.
7. i) You could draw a probability tree diagram for Jenny where the first set of branches represents the 1st attempt at the high jump and the second set of branches represents the 2nd attempt. Then you see that there are three ways that Jenny can succeed:

1. Succeed on first attempt; succeed on second attempt.
2. Succeed on first attempt; fail on second attempt.
3. Fail on first attempt; succeed on second attempt.

What you need to do is calculate these individual probabilities then add the probabilities of 1, 2 and 3 together to get the total probability of Jenny succeeding.

ii) For this one, you could work backwards. You know that the probability he succeeds is 0.51. You also know from the previous part of the question that there are three probabilities that you need to add to get the total probability of success. Work out the three probabilities in terms of p and then add them together. The resulting expression will be equal to 0.51, then you can solve for p from there.
8. (Original post by Maths_Lover)
i) You could draw a probability tree diagram for Jenny where the first set of branches represents the 1st attempt at the high jump and the second set of branches represents the 2nd attempt. Then you see that there are three ways that Jenny can succeed:

1. Succeed on first attempt; succeed on second attempt.
2. Succeed on first attempt; fail on second attempt.
3. Fail on first attempt; succeed on second attempt.

What you need to do is calculate these individual probabilities then add the probabilities of 1, 2 and 3 together to get the total probability of Jenny succeeding.

ii) For this one, you could work backwards. You know that the probability he succeeds is 0.51. You also know from the previous part of the question that there are three probabilities that you need to add to get the total probability of success. Work out the three probabilities in terms of p and then add them together. The resulting expression will be equal to 0.51, then you can solve for p from there.
I think your scenario one doesn't exist. Once she's succeeded, she doesn't have to jump again. And so 2 doesnt exist either.

It would be 0.6 ( succeed first time) + 0.4 x 0.7 (fail first, succeed second). Which is 0.88 in total.

Not sure about second one. I do it two different methods and come out with same unrealistic quadratic.

P(success) = 0.51
P(loss) = 0.49

1-p = loss first time
1-p x 1-p = loss both times = P(loss)
1 - 2p + p^2 = 0.49
P^2 - 2p + 0.51 = 0 ???

P(success) = 0.51
p = success first time
1-p x p = success second time

p - p^2 = 1-p x p

p - p^2 + p = 0.51

rearrage p^2 - 2p + 0.51 = 0.
9. (Original post by Straight up G)
I think your scenario one doesn't exist. Once she's succeeded, she doesn't have to jump again. And so 2 doesnt exist either.

It would be 0.6 ( succeed first time) + 0.4 x 0.7 (fail first, succeed second). Which is 0.88 in total.

Not sure about second one. I do it two different methods and come out with same unrealistic quadratic.

P(success) = 0.51
P(loss) = 0.49

1-p = loss first time
1-p x 1-p = loss both times = P(loss)
1 - 2p + p^2 = 0.49
P^2 - 2p + 0.51 = 0 ???

P(success) = 0.51
p = success first time
1-p x p = success second time

p - p^2 = 1-p x p

p - p^2 + p = 0.51

rearrage p^2 - 2p + 0.51 = 0.
Oh yeah. Oops, didn't read the question properly i see it now.
10. (Original post by Maths_Lover)
Oh yeah. Oops, didn't read the question properly i see it now.
Dyou think what else I've done is correct?
11. (Original post by No!)
Jenny and Omar are each allowed two attempts at a high jump.

(i) The probability that Jenny will succeed on her first attempt is 0.6. If she fails on her first attempt, the probability that she will succeed on her second attempt is 0.7.
Calculate the probability that Jenny will succeed.

(ii) The probability that Omar will succeed on his first attempt is p. If he fails on his first attempt, the probability that he will succeed on his second attempt is also p. The probability that he succeeds is 0.51. Find p.
i)0.6+ (0.4x0.7) ii) p+( (1-p) x p) =0.51
Listen to the further maths student
12. (Original post by Straight up G)
I think your scenario one doesn't exist. Once she's succeeded, she doesn't have to jump again. And so 2 doesnt exist either.

It would be 0.6 ( succeed first time) + 0.4 x 0.7 (fail first, succeed second). Which is 0.88 in total.

Not sure about second one. I do it two different methods and come out with same unrealistic quadratic.

P(success) = 0.51
P(loss) = 0.49

1-p = loss first time
1-p x 1-p = loss both times = P(loss)
1 - 2p + p^2 = 0.49
P^2 - 2p + 0.51 = 0 ???

P(success) = 0.51
p = success first time
1-p x p = success second time

p - p^2 = 1-p x p

p - p^2 + p = 0.51

rearrage p^2 - 2p + 0.51 = 0.
For the second bit, I did it this way:

Probability that he succeeds is p(succeeds first time) + p(fails then succeeds)
= p + p(1-p)
=p + p - p^2
=2p - p^2

2p - p^2 = 51/100
200p -100p^2 = 51
100p^2 - 200p +51 = 0
(10p - 3)(10p - 17) = 0

p = 0.3
13. (Original post by Legen...dary!!)
0.6+ (0.4x0.7) ii) p+( (1-p) x p) =0.51
basically what I got.
14. That leaves the roots as

1.17
and 0.3(00000000000000004).

So 0.3 for P
15. (Original post by Maths_Lover)
For the second bit, I did it this way:

Probability that he succeeds is p(succeeds first time) + p(fails then succeeds)
= p + p(1-p)
=p + p - p^2
=2p - p^2

2p - p^2 = 51/100
200p -100p^2 = 51
100p^2 - 200p +51 = 0
(10p - 3)(10p - 17) = 0

p = 0.3

Well I cheated and used a quadratic solver off the internet, but yeah, same method. I wish I'd thought of doing the quadratic that way. S1 is such a *****
16. (Original post by Straight up G)
Dyou think what else I've done is correct?
I agreed with the first part of the question that P(Jenny succeeds) = 0.88
I didn't quite get what you did for the second bit though, but I have come up with a way of doing it (see above).

17. (Original post by Straight up G)
basically what I got.
Well done G!
18. (Original post by Straight up G)
That leaves the roots as

1.17
and 0.3(00000000000000004).

So 0.3 for P
??

10p - 3 = 0
p = 3/10 = 0.3

10p - 17 = 0
p = 17/10 = 1.7 (not this root as it is more than 1, which is impossible for a probability)

Therefore p = 0.3
19. (Original post by Maths_Lover)
??

10p - 3 = 0
p = 3/10 = 0.3

10p - 17 = 0
p = 17/10 = 1.7 (not this root as it is more than 1, which is impossible for a probability)

Therefore p = 0.3
Aha, I used a quadratic solver, which gave the second root as 0.3000000000004, for some reason. Anyway, its all good.
20. (Original post by Legen...dary!!)
Well done G!
Haha
It's a big deal for me, I'm taking this exam in May.

Not that big a deal, but still happy I got it

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