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- 2013-9-12
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9#
发表于 2013-4-28 07:10
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dsjn358 wrote:
EddieChen wrote:
Yes, yours is one of the right ways, but costs you more time.
To be simple:
P(~A) = 0.4 and P(~B) = 0.65
Therefore, P(~A and ~B) = 0.4 * 0.65 = 0.26
By the way, you made two typos: 1) 0.25 but not 26 on 2nd line, and, 2) 0.6 + 0.35 instead of 0.6 + 0.36 on 5th line.
dsjn358 wrote:
I’m not sure if this is right but:
I gota answer C- 0.25
P(A: auto 5%) = 0.6
P(B: cable10%)=0.35
P(A: auto 5% or B: cable10%) = 0.6 + 0.36 - (0.6 x 0.35) = 0.95 - 0.21 = 0.74
P(NOT A: auto 5% or B: cable10%) = 1-0.74 = 0.26
is this the right way to think of this problem???
[snip]
thanks mate, the other way is much easier!
oops yes my bad,
answer C = 0.26
and yes, P(A: auto 5% or B: cable10%)= 0.6+0.35- (0.6 x 0.35) = 0.95 - 0.21 = 0.74
Isn’t it incorrect to think that there is a joint probability there? |
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