math help: Statistics (probability) - Help.com

Statistics (probability)

If given:

A∩C = .05
A∩C’ = .2
A’∩C = .03
A’∩C’ = .72

How can I find P(A) and P(C)?

This open post was written 3 weeks, 4 days ago | V/U/S: 72, 6, 1 | Edit Post | Leave a reply | Report Post


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Since writing this post pyronav may have helped people, but has not within the last 4 days. pyronav is a verified member, has been around for 3 weeks, 4 days and has 1 posts and 2 replies to their name.

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pyronav edited this post 3 weeks, 4 days ago. Read the previous text »

If given:

A∩
C = .05
A∩
C’ = .2
A’∩
C = .03
A’∩
C’ = .72

How can I find P(A) and P(C)?

sahaven offline Verified User (3 weeks, 6 days) Shouts: 97 #
An Unknown Location | 3 weeks, 4 days ago (9 minutes after post)

P(A) = 0.25
P(C) = 0.08

im 90% sure thats right lol
still got doubts…
but basically what i did was A + A’ = 1

and use that relation :)

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pyronav offline Verified User (3 weeks, 4 days) Shouts: 0 #
An Unknown Location | 3 weeks, 4 days ago (12 minutes after post)

sahaven wrote:
P(A) = 0.25
P(C) = 0.08

im 90% sure thats right lol
still got doubts…
but basically what i did was A + A’ = 1

and use that relation :)

Can you show me some steps please?

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sahaven offline Verified User (3 weeks, 6 days) Shouts: 97 #
An Unknown Location | 3 weeks, 4 days ago (21 minutes after post)

ohok
take the ‘An’ together i.e AnC and AnC’ ..use the relation that P(A) +P( A’) = 1 ( hope you know this relation from before)

this makes P(An) = P(AnC)+P(AnC’) = .05+0.2 = 0.25
also you notice that P(An’) = P(An’C) + P(An’C') = .03+0.72 = 0.75
which makes P(An)+P(An’) = 1 and hence follows the rule which i wrote in the first line.

similarly,
P(C) = P(AnC) + P(An’C) = .05+ .03 = 0.08

[ P(C’) = P(AnC’) + P(An’C') = 0.2+.72 = 0.92 and P(C)+P(C’) = .08+.92 = 1 , which also follows the rule]

hope this is understandable.

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pyronav offline Verified User (3 weeks, 4 days) Shouts: 0 #
An Unknown Location | 3 weeks, 4 days ago (23 minutes after post)

Perfect! Thank you. ^_^

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sahaven offline Verified User (3 weeks, 6 days) Shouts: 97 #
An Unknown Location | 3 weeks, 4 days ago (28 minutes after post)

glad to help :)

wish someone could help with my probability class too lol >.

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