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If A and B are two events and P(A) = 0.6, P(B) = 0.3 and P(A U B)=0.8, find:

(a) P( A n B)

(b) P( A' n B)

(c) P (A n B')

(d) P (A' n B')

(e) P (A U B')

(f) P (A' U B)

Please explain!

(a) P( A n B)

(b) P( A' n B)

(c) P (A n B')

(d) P (A' n B')

(e) P (A U B')

(f) P (A' U B)

Please explain!

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#4

If A and B are two events and P(A) = 0.6, P(B) = 0.3 and P(A U B)=0.8, find:

(a) P( A n B)

(b) P( A' n B)

(c) P (A n B')

(d) P (A' n B')

(e) P (A U B')

(f) P (A' U B)

a) probability of A and B = 0.6 x 0.3

b) probability of B but not A = 0.4 (1-0.6) x 0.3

c) probability of A but not B = 0.6 x 0.7

d) probability of not A & not B = 0.4 x 0.7

e) probability of A or Not B = 0.6 + 0.7

f) probability of Not A or B = 0.4 + 0.3

(a) P( A n B)

(b) P( A' n B)

(c) P (A n B')

(d) P (A' n B')

(e) P (A U B')

(f) P (A' U B)

a) probability of A and B = 0.6 x 0.3

b) probability of B but not A = 0.4 (1-0.6) x 0.3

c) probability of A but not B = 0.6 x 0.7

d) probability of not A & not B = 0.4 x 0.7

e) probability of A or Not B = 0.6 + 0.7

f) probability of Not A or B = 0.4 + 0.3

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#6

mmh thats odd. The rules are correct, maybe i did something wrong...

Is "If A and B are two events and P(A) = 0.6, P(B) = 0.3 and P(A U B)=0.8, find" correct"?

P(A U B)=0.8 - means probability of A union B ~ P(A) or P(B) which is P(A) + P(B)

Is "If A and B are two events and P(A) = 0.6, P(B) = 0.3 and P(A U B)=0.8, find" correct"?

P(A U B)=0.8 - means probability of A union B ~ P(A) or P(B) which is P(A) + P(B)

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#7

(Original post by

If A and B are two events and P(A) = 0.6, P(B) = 0.3 and P(A U B)=0.8, find:

(a) P( A n B)

(b) P( A' n B)

(c) P (A n B')

(d) P (A' n B')

(e) P (A U B')

(f) P (A' U B)

a) probability of A and B = 0.6 x 0.3

b) probability of B but not A = 0.4 (1-0.6) x 0.3

c) probability of A but not B = 0.6 x 0.7

d) probability of not A & not B = 0.4 x 0.7

e) probability of A or Not B = 0.6 + 0.7

f) probability of Not A or B = 0.4 + 0.3

**nas7232**)If A and B are two events and P(A) = 0.6, P(B) = 0.3 and P(A U B)=0.8, find:

(a) P( A n B)

(b) P( A' n B)

(c) P (A n B')

(d) P (A' n B')

(e) P (A U B')

(f) P (A' U B)

a) probability of A and B = 0.6 x 0.3

b) probability of B but not A = 0.4 (1-0.6) x 0.3

c) probability of A but not B = 0.6 x 0.7

d) probability of not A & not B = 0.4 x 0.7

e) probability of A or Not B = 0.6 + 0.7

f) probability of Not A or B = 0.4 + 0.3

0.8 = 0.6 +0.3 - p(A n B)

Therefore P(A n B) = 0.1

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#9

(Original post by

Isn't it P( A n B) which means A and B ~ A X B

**nas7232**)Isn't it P( A n B) which means A and B ~ A X B

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#10

Yeah i got a B for s1

Repeating this year to try n get near 100%

Haven't done stats in a while, i scheduled revision for april ish.

Whats the "lol" for, that's not very nice issit.

..looked in my edexcel book; got signs mixed up. Ignore the answers

Repeating this year to try n get near 100%

Haven't done stats in a while, i scheduled revision for april ish.

Whats the "lol" for, that's not very nice issit.

..looked in my edexcel book; got signs mixed up. Ignore the answers

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#12

Use a Venn diagram.

P(A n B) = P(A) + P(B) - P(A u B) = 0.1

P(A' n B) = P(B) - P(A n B) = 0.2

P(A n B') = P(A) - P(A n B) = 0.5

P(A' n B') = 1 - P(A u B) = 0.2

P(A u B') = 1 - P(A' n B) = 0.8

P(A' u B) = 1 - P(A n B') = 0.5

P(A n B) = P(A) + P(B) - P(A u B) = 0.1

P(A' n B) = P(B) - P(A n B) = 0.2

P(A n B') = P(A) - P(A n B) = 0.5

P(A' n B') = 1 - P(A u B) = 0.2

P(A u B') = 1 - P(A' n B) = 0.8

P(A' u B) = 1 - P(A n B') = 0.5

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