# a2 probability

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#1
hi, could someone give me a hand with this please? i'm just not that sure where to start

given that events A and B are independent, find p(anb)
P(a)= 0.3 and p(aub)=0.7

thanks
0
4 weeks ago
#2
p(aub) = P(a) + P(b) - P(anb)
Hence we have 0.7 = 0.3 + P(b) - P(anb), and P(b) - P(anb) = 0.4

Since they are independent:
P(anb) = P(a) x P(b), so P(b) = P(anb) / 0.3

At this point you can substitute in to our first equation (P(b) - P(anb) = 0.4) and rearrange to find P(anb) 0
#3
ahhh i get it now, thank you!
i got 0.12 does that sound about right?
0
4 weeks ago
#4
(Original post by marnieeeee)
ahhh i get it now, thank you!
i got 0.12 does that sound about right?
I got 6/35 but I think I see how you'd get 0.12 - I think you've used 0.4 as P(B), rather than 0.4 + P(AnB).

You essentially should end up with
P(b) - P(anb) = 0.4
P(b) = P(anb) / 0.3
Hence (P(anb) / 0.3) - P(anb) = 0.4
Multiply through by 0.3:
P(anb) - 0.3P(anb) = 0.12
0.7P(anb) = 0.12
2
#5
(Original post by Interea)
I got 6/35 but I think I see how you'd get 0.12 - I think you've used 0.4 as P(B), rather than 0.4 + P(AnB).

You essentially should end up with
P(b) - P(anb) = 0.4
P(b) = P(anb) / 0.3
Hence (P(anb) / 0.3) - P(anb) = 0.4
Multiply through by 0.3:
P(anb) - 0.3P(anb) = 0.12
0.7P(anb) = 0.12
ohhh i see, sorry that was such a silly mistake! thank you for all your help 0
4 weeks ago
#6
(Original post by marnieeeee)
ohhh i see, sorry that was such a silly mistake! thank you for all your help no problem, I made the same mistake at first too 0
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