(B) The total number of past working days is $200$.Let $E$ be the event that tomorrow's output will have more than $13$ defective parts.From the given

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(B) The total number of past working days is $200$.Let $E$ be the event that tomorrow's output will have more than $13$ defective parts.From the given table,we observe that the maximum number of defective parts recorded is $13$.Therefore,the number of days on which more than $13$ defective parts were produced is $0$.The probability $P(E)$ is calculated as:$P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$$P(E) = \frac{0}{200} = 0$.

Class 9 Mathematics Probability Mix Examples - Probability

Medium

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

Number 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.

A
$1$
B
$0$
C
$\frac{1}{200}$
D
$\frac{13}{200}$

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Class 9

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Probability

Subtopic

Mix Examples - Probability

$A$ die is thrown $1000$ times and the outcomes were recorded as follows:

Outcome $1$ $2$ $3$ $4$ $5$ $6$

Frequency $180$ $150$ $160$ $170$ $150$ $190$

If the die is thrown once more,then the probability that it shows $5$ is:

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The probability of occurrence of event $A$ is $0.43$. The probability of event $A$ not occurring is .........

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The maximum value of the probability of any event is ...........

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Bulbs are packed in cartons,each containing $40$ bulbs. Seven hundred cartons were examined for defective bulbs,and the results are given in the following table:

Number of defective bulbs $0$ $1$ $2$ $3$ $4$ $5$ $6$ more than $6$

Frequency $400$ $180$ $48$ $41$ $18$ $8$ $3$ $2$

One carton was selected at random. What is the probability that it has less than $4$ defective bulbs?

Medium

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Here is an extract from a mortality table.

Age (in years) Number of persons surviving out of a sample of one million

$60$ $16090$

$61$ $11490$

$62$ $8012$

$63$ $5448$

$64$ $3607$

$65$ $2320$

What is the probability that a person aged $61$ will live for $4$ years?

Medium

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Number of defective parts	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$	$13$
Days	$50$	$32$	$22$	$18$	$12$	$12$	$10$	$10$	$10$	$8$	$6$	$6$	$2$	$2$

Outcome	$1$	$2$	$3$	$4$	$5$	$6$
Frequency	$180$	$150$	$160$	$170$	$150$	$190$

Number of defective bulbs	$0$	$1$	$2$	$3$	$4$	$5$	$6$	more than $6$
Frequency	$400$	$180$	$48$	$41$	$18$	$8$	$3$	$2$

Age (in years)	Number of persons surviving out of a sample of one million
$60$	$16090$
$61$	$11490$
$62$	$8012$
$63$	$5448$
$64$	$3607$
$65$	$2320$

Similar Questions

$A$ die is thrown $1000$ times and the outcomes were recorded as follows: Outcome $1$ $2$ $3$ $4$ $5$ $6$ Frequency $180$ $150$ $160$ $170$ $150$ $190$ If the die is thrown once more,then the probability that it shows $5$ is:

The probability of occurrence of event $A$ is $0.43$. The probability of event $A$ not occurring is .........

The maximum value of the probability of any event is ...........

Here is an extract from a mortality table. Age (in years) Number of persons surviving out of a sample of one million $60$ $16090$ $61$ $11490$ $62$ $8012$ $63$ $5448$ $64$ $3607$ $65$ $2320$ What is the probability that a person aged $61$ will live for $4$ years?

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$A$ die is thrown $1000$ times and the outcomes were recorded as follows:

Outcome $1$ $2$ $3$ $4$ $5$ $6$

Frequency $180$ $150$ $160$ $170$ $150$ $190$

If the die is thrown once more,then the probability that it shows $5$ is:

Here is an extract from a mortality table.

Age (in years) Number of persons surviving out of a sample of one million

$60$ $16090$

$61$ $11490$

$62$ $8012$

$63$ $5448$

$64$ $3607$

$65$ $2320$

What is the probability that a person aged $61$ will live for $4$ years?