Probability Test

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Probability Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

Two coins are tossed $$1000$$ times and the outcomes are recorded as below:
No of heads
$$2$$
$$1$$
$$0$$
Frequency
$$200$$
$$550$$
$$250$$
Based on this information, the probability for at most one head is $$\dfrac{a}{b}$$, where $$(a,b)=1$$. Then $$\dfrac{a}{b}$$ is

A
$$\dfrac{1}{5}$$

B
$$\dfrac{1}{4}$$

C
$$\dfrac{4}{5}$$

D
$$\dfrac{3}{4}$$

Solution

Total number of outcomes $$= 1000$$
Number of favourable outcomes $$= 800$$
Probability = $$\dfrac{800}{1000}$$ = $$\dfrac{4}{5}$$

Question 2

1 / -0

There are $$500$$ packets in a large box and each packet contains $$4$$ electronic devices in it. On testing, at the time of packing, it was noted that there are some faulty pieces in the packets. The data is as below :

No. of faulty devices in a packet	Number of packets
$$0$$	$$300$$
$$1$$	$$100$$
$$2$$	$$50$$
$$3$$	$$30$$
$$4$$	$$20$$
Total number of packets	$$500$$

If one packet is drawn from the box, what is the probability that all the four devices in the packet are without any fault?

$$0.5$$

$$0.6$$

$$0.8$$

$$0.9$$

Solution

When the packet has all the four devices without fault, it means the number of faulty devices in the packet is $$0$$. Number of chances which are favourable to $$0$$ are $$300$$ as given in the table above.
Thus, the probability of packet containing all the four devices without any fault $$=$$ $$ \displaystyle \frac{300}{500} $$ $$=$$ $$ \displaystyle \frac{3}{5} $$ $$= 0.6$$

Question 3
1 / -0

Fifty seeds were selected at random from each of $$5$$ bags $$A, B, C, D, E$$ of seeds, and were kept under standardised conditions equally favourable to germination. After $$20$$ days, the number of seeds which had germinated in each collection were counted and recorded as follow :
Bag $$A$$ $$B$$ $$C$$ $$D$$ $$E$$
Number of seeds
germinated
$$40$$ $$48$$ $$42$$ $$39$$ $$41$$
What is the probability of germination of
$$(i)$$ more than $$40$$ seeds in a bag?
$$(ii)$$ $$49$$ seeds in a bag?
$$(iii)$$ more than $$35$$ seeds in a bag?

A
$$(i)\, 0.690$$
$$(ii)\, 0.09$$
$$(iii)\, 1$$

B
$$(i)\, 0.80$$
$$(ii)\, 0.006$$
$$(iii)\, 1$$

C
$$(i)\, 0.70$$
$$(ii)\, 0.001$$
$$(iii)\, 1$$

D
$$(i)\, 0.60$$
$$(ii)\, 0$$
$$(iii)\, 1$$

Solution

(i) Number of bags in which more than $$40$$ seeds out of $$50$$ seeds germinated $$= 3$$
So, $$P$$ (more than $$40$$ seeds in a bag germinated) $$=$$ $$ \displaystyle \frac{3}{5} $$ $$= 0.60$$

(ii)Number of bags in which $$49 $$ seeds germinated $$= 0$$
So, $$P$$ ($$49$$ seeds germinated in bag) $$=$$ $$ \displaystyle \frac{0}{5} $$ $$= 0$$

(iii) Number of bags in which more than $$35$$ seeds out $$50$$ seeds germinated $$= 5$$
Total number of bags$$ = 5$$
So, $$P $$(more than $$35$$ seeds in a bag germinated) $$=$$ $$ \displaystyle \frac{5}{5} $$ $$= 1$$
Question 4
1 / -0

A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table show the result of $$1000$$ cases :
Distance in $$Km$$ Frequency
Less than $$4000$$ $$20$$
$$4000$$ to $$9000$$ $$210$$
$$9000$$ to $$14000$$ $$325$$
More than $$14000$$ $$445$$
If you buy a tyre of this company what is the probability that it will need to be replaced after it has covered somewhere between $$4000\, km$$ and $$14000\, km$$?

A
$$0.65$$

B
$$0.625$$

C
$$0.125$$

D
None of these

Solution

Probability that it will need to be replaced before it has covered $$ 4000 $$ km $$ = \dfrac {20}{1000} = 0.02 $$

Probability that it will last more than $$ 9000 km = \dfrac { 325 + 445 } {1000} = \dfrac {770}{1000} = 0.77 $$
Probability that it will need to be replaced after it has covered somewhere between $$ 4000 $$ km and $$ 14000 $$ km = $$ \dfrac { 210 + 325 } {1000} = \dfrac {535}{1000} = 0.535 $$
Question 5
1 / -0

A survey was conducted by car manufacturing company in a metropolitan city on $$1000$$ persons having monthly income from Rs. $$30,001$$ to Rs. $$50,000$$. The data about the number of persons in various categories is as under:

Monthly income

Number of Cars

(in rupees)

$$1$$

$$2$$

More than $$2$$

$$30,001-40,000$$

$$400$$

$$50$$

$$25$$

$$40,001-50,000$$

$$100$$

$$300$$

$$125$$

Find the probability that a person selected at random in the income slab $$40,001-50,000$$ have more than $$2$$ cars.

A
$$0.125$$

B
$$0.225$$

C
$$0.325$$

D
None of thesee

Solution

A survey was conducted by car manufacturing company in a metropolitan city on $$1000$$ persons having monthly income from Rs. $$30,001$$ to Rs. $$50,000$$
In the income slab $$40,001-50,000$$ having more than $$2$$ cars in the company is $$125$$
Then the probability that a person $$=\dfrac{125}{1000}=0.125$$
Question 6
1 / -0

A couple of dice rolled. What is the empirical probability of getting the sum as $$5$$?

A
$$\dfrac{3}{2}$$

B
$$\dfrac{1}{9}$$

C
$$\dfrac{5}{2}$$

D
$$\dfrac{7}{2}$$

Solution

Refer below diagram for sample space of two dice rolled to get sum of $$5$$

When a pair of dice rolled possibilities for getting $$5$$ is $$(1, 4), (2, 3), (3, 2), (4, 1)$$

Total number events occurred when we roll the dice $$= 36$$

So, the empirical probability formula is as follows,

$$P(E) =$$ $$\dfrac{4}{36}= \dfrac{1}{9}$$
Question 7
1 / -0

The record of a weather station shows that out of the past $$250$$ consecutive days, its weather forecasts were correct $$175$$ times. What is the probability that on a given day it was correct?

A
$$0.9$$

B
$$0.8$$

C
$$0.7$$

D
$$0.6$$

Solution

Total no. of days for which record of weather is given $$=250$$
No. of days for which forecasts were correct $$= 175$$
$$\therefore $$ favorable days $$= 175$$
$$\therefore $$ probability $$= \displaystyle \frac {175}{250} = 0.7 $$
Question 8
1 / -0

A box has tokens numbered $$3$$ to $$100$$. If a token is taken out at random the chance that the number is divisible by $$7$$ is

A
$$\dfrac{1}{4}$$

B
$$\dfrac{1}{5}$$

C
$$\dfrac{1}{6}$$

D
$$\dfrac{1}{7}$$

Solution

The tokens are numbered from $$3$$ to $$100$$
There are $$98$$ numbers from $$3$$ to $$100$$

In this range the numbers divisible by $$7$$ are $$7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91$$ and $$98$$
The number of numbers divisible by $$7$$ is $$14$$

$$P$$(number divisible by $$7$$)$$=\displaystyle \frac{14}{98}=\frac{1}{7}$$

Hence, the required probability is $$\dfrac{1}{7}$$
Question 9
1 / -0

You have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, if the probability of getting a non-blue sector is $$\dfrac{4}{m}$$. Then, the value of $$m$$ is

A
$$5$$

B
$$4$$

C
$$3$$

D
$$1$$

Solution

Out of 5 sectors, the pointer can stop at any of the sectors in 5 ways.

$$\therefore $$ Total number of outcomes = $$5$$

There are 4 non-blue sectors in the spinning wheel, out of which one can be obtained in 4 ways.

So, the required probability = $$\dfrac{4}{5}$$

Hence the value of $$m$$ is $$5$$
Question 10
1 / -0

A box contains $$32$$ coloured marbles. $$8$$ of them are red marbles and the rest are either blue or green marbles. A marble is drawn at random. Calculate the probability of drawing a marble which is not red in colour.

A
$$\dfrac {2}{3}$$

B
$$\dfrac {5}{8}$$

C
$$\dfrac {3}{4}$$

D
$$\dfrac {7}{16}$$

Solution

The marbles not in red color are $$32-8=24$$
Then probability of not drawing red marbles is,
$$P(E)=\dfrac{24}{32}\\\\\ \ \ \ \ \ \ \ =\dfrac{3}{4}$$

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Distance in $$Km$$	Frequency
Less than $$4000$$	$$20$$
$$4000$$ to $$9000$$	$$210$$
$$9000$$ to $$14000$$	$$325$$
More than $$14000$$	$$445$$

Monthly income	Number of Cars
(in rupees)	$$1$$	$$2$$	More than $$2$$
$$30,001-40,000$$	$$400$$	$$50$$	$$25$$
$$40,001-50,000$$	$$100$$	$$300$$	$$125$$

No of heads	$$2$$	$$1$$	$$0$$
Frequency	$$200$$	$$550$$	$$250$$

Probability Test - 17

Result

Probability Test - 17

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SHARING IS CARING

Question 1

$$\dfrac{1}{5}$$

$$\dfrac{1}{4}$$

$$\dfrac{4}{5}$$

$$\dfrac{3}{4}$$

Solution

Question 2

$$0.5$$

$$0.6$$

$$0.8$$

$$0.9$$

Solution

Question 3

$$(i)\, 0.690$$$$(ii)\, 0.09$$$$(iii)\, 1$$

$$(i)\, 0.80$$$$(ii)\, 0.006$$$$(iii)\, 1$$

$$(i)\, 0.70$$$$(ii)\, 0.001$$$$(iii)\, 1$$

$$(i)\, 0.60$$$$(ii)\, 0$$$$(iii)\, 1$$

Solution

Question 4

$$0.65$$

$$0.625$$

$$0.125$$

None of these

Solution

Question 5

$$0.125$$

$$0.225$$

$$0.325$$

None of thesee

Solution

Question 6

$$\dfrac{3}{2}$$

$$\dfrac{1}{9}$$

$$\dfrac{5}{2}$$

$$\dfrac{7}{2}$$

Solution

Question 7

$$0.9$$

$$0.8$$

$$0.7$$

$$0.6$$

Solution

Question 8

$$\dfrac{1}{4}$$

$$\dfrac{1}{5}$$

$$\dfrac{1}{6}$$

$$\dfrac{1}{7}$$

Solution

Question 9

$$5$$

$$4$$

$$3$$

$$1$$

Solution

Question 10

$$\dfrac {2}{3}$$

$$\dfrac {5}{8}$$

$$\dfrac {3}{4}$$

$$\dfrac {7}{16}$$

Solution

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$$(i)\, 0.690$$
$$(ii)\, 0.09$$
$$(iii)\, 1$$

$$(i)\, 0.80$$
$$(ii)\, 0.006$$
$$(iii)\, 1$$

$$(i)\, 0.70$$
$$(ii)\, 0.001$$
$$(iii)\, 1$$

$$(i)\, 0.60$$
$$(ii)\, 0$$
$$(iii)\, 1$$