2 years ago

Given P(A) = 0.46, P(B) = 0.35 and P(B|A) = 0.45, find the value of

Mathematics

1 answer:

WINSTONCH [101]2 years ago

3 0

P(B|A) = P(A and B)/P(A)
So to find P(A and B) just multiply P(B|A) by P(A)

P(A and B) = 0.45*0.46 = 0.207

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A store gives a customer 20% off if $10 or more amount of product is purchased. Otherwise the customer pays the usual price at c

Nadya [2.5K]

Answer:

The expected payment by the customer at the checkout is $9.

Step-by-step explanation:

The amount of the product is given as

$f(x)=\left \{ {{\frac{50}{x^3} \,\,\,\,\, x\geq 5 } \atop {0}} \right.$

Now the expected payment is given as

$EP=\int\limits^{10}_{5} {x f(x)} \, dx +\int\limits^{\infty}_{10} {0.8x f(x)} \, dx$

Here 0.8 x is used in the second integral because of the discount of 20% i.e. the expected price is 80% of the value such that

$\\EP=\int\limits^{10}_{5} {x \frac{50}{x^3}} \, dx +\int\limits^{\infty}_{10} {0.8x \frac{50}{x^3}} \, dx\\\\EP=\int\limits^{10}_{5} {\frac{50}{x^2}} \, dx +\int\limits^{\infty}_{10} {\frac{40}{x^2}} \, dx\\EP=[\frac{50}{-x}]_5^{10} +[\frac{40}{-x}]_{10}^{\infty} \\EP=[\frac{-50}{10}+\frac{50}{5}] +[\frac{-40}{\infty}+\frac{40}{10}]\\\\EP=-5+10+0+4\\EP=9$