English, 28.10.2019 14:46 nelgelinagudo

Find the probability that the second battery selected is not defective.

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Answer from: Laurenjayshree

2/3 or 0.6667 or 66.67% (There are different ways to write the probability either fraction, decimal or percent)

Step-by-step explanation:

In computation of probability, dont forget the keyword of combinatorics (permutation and combination).

In the given problem, since two batteries selected randomly, it was combination.

The formula for probability is P(E) = E/S

where:

E= No. of elements of events

E= No. of elements of eventsS= No. of elements of sample

To get the total number of elements for sample, use combinations since the worded problem will be selecting 2 batteries. Take note: Since it is selected randomly, you will get different combinations of batteries.

3 Batteries are not defective, and 1 is defective

The condition in the given problem is selecting 2 random batteries but not defective.

Let n is number of batteries, and r is number of batteries selected

For not defective batteries n=3, and r=2, and for defective batteries n=1 and r=0 since there is no selection for defective.

Combination= nCr = n!/(n-r)!r!

Therefore, S= nCr of not defective X nCr of deffective

S= (3C2)(1C0)

S= 3

To calculate the probability using the formula given above.

P(E) = E/S, where E=2 and S=3

P(E) = E/S,

P(E) = 2/3

Therefore, the probability of selecting 2 randomly not defective is 2/3 (fraction), 0.6667 (decimal form), or 66.67% (percentage form).

#BrainlyFast

Read more about combinatorics at

Answer from: rhaineandreirefuerzo

D. 1/2

Step-by-step explanation:

What is asked?

The probability that the second battery selected is not defective given that the first was not defective.

What are given?

Four batteriesN = not defectiveD = defective

Condition:

one of the four batteries is defectiveboth the batteries are not defective

Formula:

First we find all the possible outcomes or the sample space of the problem. With that, we use the combination formula.

$_n C_r = \frac{n! }{(n-r)! r! }$

Since we are to choose 2 out of fourThe sample space is given by

$_n C_r = \frac{n! }{(n-r)! r! } \\ _4 C_2 = \frac{4! }{(4-2)! 2! } \\ _4 C_2 = \frac{4! }{2! 2! } \\ _4 C_2 = 6$

Listing the sample space N = a, b, c and D=x

A = {ax, bx, cx, ab, ac, bc}

Seeing the list of sample space, we need to find the element(s) that suits the second condition. The condition that suits the condition is only ab, ac, and bc. So we have 3.

$P(y) = \frac{3}{6} = \frac{1}{2}$

There is 1/2 probability that the selected is not defective given that the other one is not defective.

What is combination? Read here:

#BrainlyFast

Answer from: elishakim80

The answer would be $\frac{2}{3}$

Step-by-step explanation:

= 4 batteries - 1 (defective) = 3

= Two are selected random for use on a particular day

= 2/3

Since we are given that the first battery was not defective, and assuming

the we are sampling without replacement, we reduce the sample space to

three batteries, one of which is defective and two of which are not

defective. So the probability of the second one not being defective

is now 2/3. #BrainlyFast

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