Over the past 200 working days, the number of defective parts produced by a machine is given in the following table:

No of defective parts
0
1
2
3
4
5
6
7
8
9
10
11
12
13

Days
50
32
22
18
12
12
10
10
10
8
6
6
2
2

Determine the probability that tomorrow's output will have at least one defective part.

0.25

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0.50

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0.66

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0.75

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Solution

The correct option is C $0.75$
The no. of possible outcomes $= 14$ ...[starting from defect count =0 to defect count =13]

The no. of days required $= 50$ to $2$ days

So, $P (no defect) = (\frac{50}{200}) = 0.25$

The probability that there will be at least 1 defect or all 13 defects

$= 1 - P (no defect) = 1 - 0.25 = 0.75$

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Similar questions

Over the past 200 working days, the number of defective parts produced by a machine is given in the following table:

No of defective parts	0	1	2	3	4	5	6	7	8	9	10	11	12	13
Days	50	32	22	18	12	12	10	10	10	8	6	6	2	2

Determine the probability that tomorrow's output will have more than 13 defective parts.

Over the past 100 days, the number of defective bulbs produced by a machine is given by the following.

No. of defective bulbs	0	1	2	3
Days	20	40	12	28

The probability that tomorrow's output will have no defect.

Q. Question 19
Over the past 200 working days, the number of defective parts produced by a machine is given in the following table.

\begin{matrix} N u m b e r o f & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 d e f e c t i v e p a r t s D a y s & 50 & 32 & 22 & 18 & 12 & 12 & 10 & 10 & 10 & 10 & 8 & 6 & 6 & 2 \end{matrix}

Determine the probability that tomorrow’s output will have:
(i) No defective part.
(ii) Atleast one defective part.
(iii) Not more than 5 defective parts.
(iv) More than 13 defective parts.

Over the past 200 working days, the number of defective parts produced by a machine is given in the following table.

$\begin{matrix} Number of & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 defective parts Days & 50 & 32 & 22 & 18 & 12 & 12 & 10 & 10 & 10 & 10 & 8 & 6 & 6 & 2 \end{matrix}$

Determine the probability that tomorrow’s output will have:
(i) No defective part.
(ii) Atleast one defective part.
(iii) Not more than 5 defective parts.
(iv) More than 13 defective parts.

Q. Suppose a machine produces metal parts that contain some defective parts with probability

0.05

. How many parts should be produced in order that probability of at least one part being defective is

1 / 2

or more? (Given

{log}_{10} 95 = 1.977

and

{log}_{10} 2 = 0.3)

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Over the past 200 working days, the number of defective parts produced by a machine is given in the following table:No of defective parts012345678910111213Days50322218121210101086622Determine the probability that tomorrow's output will have at least one defective part.

The correct option is C 0.75The no. of possible outcomes =14 ...[starting from defect count =0 to defect count =13]The no. of days required =50 to 2 daysSo, P(no defect)=(50200)=0.25 The probability that there will be at least 1 defect or all 13 defects=1−P(no defect)=1−0.25=0.75