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AQA S1 binomial distribution help please

Hi all,

I don't understand why the probability of success in 4(b)(iii) used is 0.15? Why is it not (0.15)(0.80)?

My calculation:
Mean = np
n = 80
p = (0.15)(0.80)
Mean = (80)(0.15)(0.80)
Mean = 9.6

Correct calculation:
Mean = no problem
n = 80
p = (0.15)
Mean = (80)(0.15)
Mean = 12

Thank you very much!
Original post by londoncricket
Hi all,

I don't understand why the probability of success in 4(b)(iii) used is 0.15? Why is it not (0.15)(0.80)?

My calculation:
Mean = np
n = 80
p = (0.15)(0.80)
Mean = (80)(0.15)(0.80)
Mean = 9.6

Correct calculation:
Mean = no problem
n = 80
p = (0.15)
Mean = (80)(0.15)
Mean = 12

Thank you very much!


Question is asking the probabilty that she uses her second barrel, NOT that she uses her second barrel and hits the target. We're not interested in whether she hits the target or not with the second barrel, only that she uses it.
Original post by londoncricket
Hi all,

I don't understand why the probability of success in 4(b)(iii) used is 0.15? Why is it not (0.15)(0.80)?

My calculation:
Mean = np
n = 80
p = (0.15)(0.80)
Mean = (80)(0.15)(0.80)
Mean = 9.6

Correct calculation:
Mean = no problem
n = 80
p = (0.15)
Mean = (80)(0.15)
Mean = 12

Thank you very much!


We want to know how many times she fires the second barrel. This is the same as saying that we want the number of times she doesn't hit the first shot. P(First shot hits) = 0.85, so P(First shot misses) = 0.15.

Mean = np
= 80*0.15
= 12

Hope this helps.
Original post by britishtf2
We want to know how many times she fires the second barrel. This is the same as saying that we want the number of times she doesn't hit the first shot. P(First shot hits) = 0.85, so P(First shot misses) = 0.15.

Mean = np
= 80*0.15
= 12

Hope this helps.


Ah I understand!

Thank you so much for this!
Original post by londoncricket
Ah I understand!

Thank you so much for this!


No problem. DM with any questions you might have, and I'll try to help :smile:
Reply 5
In regards to the above, do not 'dm' people with requests for help, the maths forum is here for that. :smile:

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