Question

3 batteries are chosen at random from 15 batteries from which 5 are defective. Find the probability that:

A) none of the three are defective
B) exaxtly one is defective
C) 2 defective and one nondefective
D)at least one is nondefective.

Please type the solution. Im struggling~~ :( Please help meeeee~ THANK YOUUU~

Answer 1

Answer on Question #64948 – Math – Statistics and Probability

Question

3 batteries are chosen at random from 15 batteries from which 5 are defective. Find the probability that:

A) none of the three are defective;

B) exactly one is defective;

C) 2 defective and one nondefective;

D) at least one is nondefective.

Solution

A) Let's take the first battery from 15 batteries and denote the event by A. The probability of chosen battery not being defective will be 10/15.

Let's take the next battery from other 14 batteries and denote the event by B. The probability of chosen battery not being defective will be 9/14.

Let's take the last battery from remaining 13 batteries and denote the event by C. The probability of that battery not being defective will be 8/13.

The probability of none of the three being defective is calculated as the product

$$P _ {1} = P (3 \text { non defective}) = P (A, B, C) = \frac {1 0}{1 5} \cdot \frac {9}{1 4} \cdot \frac {8}{1 3} = \frac {7 2 0}{2 7 3 0} \approx 0. 2 6 4$$

B) The number of combinations of choosing 3 batteries from 15 is $C_{15}^{3}$

$C_5^1$ is the number of combinations of choosing 1 defective battery from 5 defective ones. $C_{10}^{2}$ is the number of combinations of choosing 2 good batteries from 10 good ones.

So $C_5^1 C_{10}^2$ is the total amount of combinations to choose exactly one defective and two good batteries. The probability of exactly one battery being defective will be

$$\begin{array}{l} P _ {2} = P (2 \text { non defective}, 1 \text { defective}) = \frac {C _ {5} ^ {1} C _ {1 0} ^ {2}}{C _ {1 5} ^ {3}} = \frac {\frac {5 !}{1 ! 4 !} \frac {1 0 !}{2 ! 8 !}}{\frac {1 5 !}{3 ! 1 2 !}} = \frac {\frac {5 \cdot 1 0 \cdot 9}{2}}{\frac {1 5 \cdot 1 4 \cdot 1 3}{3 \cdot 2}} \\ = \frac {3 \cdot 5 \cdot 9 \cdot 1 0}{1 3 \cdot 1 4 \cdot 1 5} = \frac {4 5}{9 1} \\ \end{array}$$

C) The arguments are similar to ones in the previous part B) and the probability of choosing 2 from 5 defective batteries and 1 from 10 good batteries will be

P_3 = P(1 \text{ nondefective}, 2 \text{ defective}) = \frac{C_5^2 C_{10}^1}{C_{15}^3} = \frac{\frac{5!}{2!} \cdot \frac{10!}{3! \cdot 12!} \cdot \frac{10!}{15!} = \frac{\frac{5 \cdot 4}{2} \cdot 10}{\frac{15 \cdot 14 \cdot 13}{3 \cdot 2}} = \frac{10 \cdot 3 \cdot 4 \cdot 5}{13 \cdot 14 \cdot 15} = \frac{20}{91}

D)

$P_4 = P(\text{at least 1 is nondefective}) = P(1 \text{ nondefective}, 2 \text{ defective}) + P(2 \text{ nondefective}, 1 \text{ defective}) + P(3 \text{ nondefective}) = P_3 + P_2 + P_1 = \frac{20}{91} + \frac{45}{91} + \frac{720}{2730} \approx 0.978.$

Answer:

A) 0.264;

B) $\frac{45}{91}$;

C) $\frac{20}{91}$;

D) 0.978.

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