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sukhopar [10]
4 years ago
12

Mrs. Jones buys two toys for her son. The probability that the first toy is defective is 1/3, and the probability that the secon

d toy is defective given that the first toy is defective is 1/5. What is the probability that both toys are defective?
Mathematics
2 answers:
Debora [2.8K]4 years ago
6 0

Answer: \frac{1}{15}

Step-by-step explanation:

Let F denote the event of first  toy is defective and S denote the second toy is defective .

Given: The probability that the first toy is defective P(F)=\frac{1}{3}

The probability that the second toy is defective given that the first toy is defective  P(S|F)=\frac{1}{5}

The formula to calculate the conditional probability is given by :-

P(S|F)=\frac{P(S\cap F)}{P(F)}\\\\\Rightarrow P(S\cap F)=P(S|F)\times P(F)\\\\\Rightarrow\ P(S\cap F)=\frac{1}{3}\times \frac{1}{5}=\frac{1}{15}

Hence, the probability that both toys are defective=\frac{1}{15}

sergeinik [125]4 years ago
3 0

Answer:

Probability=\frac{1}{15}

Step-by-step explanation:

As it is given that

Probability of toy A is defective is =P(A) = \frac{1}{3}

Probability of toy b is defective if A is defective = P (B)=\frac{1}{5}

WE have to find the P(A n B)

By the law of Probability

P(A n B) = P (A).P(B)

putting the values given to us

P(AnB)=\frac{1}{3} * \frac{1}{5}

Probability=\frac{1}{15}

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Roman55 [17]

The two parabolas intersect for

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and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

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b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

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\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

Using the result of part (a), the volume is

\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

and using the result of part (a) again, the volume is

\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

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