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Mariulka [41]
3 years ago
13

Andrew has six fewer coins than Casey. Bob has two more than twice as many coins as Andrew. Altogether, they have 544 coins. How

many coins does Bob have?
Mathematics
1 answer:
Damm [24]3 years ago
8 0
Assuming: number of coins Andrew has=a, number of coins Bob has=b, number of coins Casey has=c
Andrew has six fewer coins than Casey, so, c-6=a, or a+6=c
Bob has two more than twice as many coins as Andrew, so 2a+2=b
Together, there are 544 coins: a+b+c=544. Rewrite the equation in terms of a: a+(2a+2)+(a+6)=544. Combine like terms: 4a+8=544. Solve the equation: 4a=536, so a=134. If b=2a+2, then b=2(134)+2=268+2=270.
Bob has 270 coins.
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P(B)=P(B|A)*P(A)+P(B|A)*P(~A)

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Note: get used to working in fraction when doing probability.

(a) Find P(A|B):

By Baye's Theorem,

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(b) Find P(~A|~B)

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P(~A)=1-P(A)=399/400

P(~B)=1-P(B)=133/136

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=P(B|A)*P(A) [def. of cond. prob.]

=9/10*(1/400)

=9/4000

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=P(A)+P(B)-P(A∩B)

=1/400+51/500-9/4000

=409/4000

P(~A|~B)

=P(~A∩~B)/P(~B)

=P(~A∪B)/P(~B)

=(1-P(A∪B)/(1-P(B)) [ law of complements ]

=(3591/4000) ÷ (449/500)

=3591/3592

The results can be easily verified using a contingency table for a random sample of 4000 persons (assuming outcomes correspond exactly to probability):

===....B...~B...TOT

..A . 9 . . 1 . . 10

.~A .399 .3591 . 3990

Tot .408 .3592 . 4000

So P(A|B)=9/408=3/136

P(~A|~B)=3591/3592

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