3 Marks Question

Question 13 Marks

In a circket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that he did not hit a boundary.

Answer

Total number of ball played = 30
Number of times boundary was hit = 6
⇒ Number of times boundary was not hit = 30 - 6 = 24
Therefore,
Probability that the batsman did not hit the boundary
$=\frac{\text{No. of times boundary was not hit}}{\text{Total number of balls played}}$
$=\frac{24}{30}$
$=\frac{4}{5}=0.8$

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Question 23 Marks

On one page of a telephone directory, there are $200$ phone numbers. The frequency distribution of their unit's digits is given below:

Unit's digit	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
Frequency	$19$	$22$	$23$	$19$	$21$	$24$	$23$	$18$	$16$	$15$

One of the numbers is chosen at random from the page. What is the probability that the unit's digit of the chosen number is:

$5?$
$8?$

Answer

Total phone numbers on the directory page $= 200$

Number of numbers with units digits $5 = 24$

Let $E_1$ be the event that the units digits of selected number is $5.$
$\therefore$ Required probability $= P(E_1) =\frac{24}{200}=0.12$

Number of numbers with units digits $8 = 16$

Let $E_2$ be the event that the units digits of selected number is $8.$
$\therefore$ Required probability $= P(E_2) =\frac{16}{200}=0.08$

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Question 33 Marks

The following table shows the blood groups of $40$ students of a class.

Blood group	$A$	$B$	$O$	$AB$
Number of students	$11$	$9$	$14$	$6$

One student of the class is chosen at random. What is the probability that the chosen students blood group is:
$i. O? \ \ \ \ \ ii. AB?$

Answer

Total number of students $= 40$
$i.$ Number of students with blood group $O = 14$
Let $E_1$ be the event that the selected student's blood group is $O.$
$\therefore$ Required probability $= P(E_1) =\frac{14}{40}=0.35$
$ii.$ Number of students with blood group $AB = 6$
Let $E_2$ be the event that the selected student's blood group is $AB.$
$\therefore$ Required probability $= P(E_2) =\frac{6}{40}=0.15$

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Question 43 Marks

The table given below shows the marks obtained by 30 students in a test.

Marks (Class interval)	1 - 10	11 - 20	21 - 30	31 - 40	41 - 50
Number of students (Frequency)	7	10	6	4	3

Out of these students, one is chosen at random. What is the probability that the marks of the chosen student:

Are 30 or less?
Are 31 or more?
Lie in the interval 21 - 30?

Answer

Total number of students = 30

Probability that the marks of the chosen student are 30 or less $=\frac{7+10+6}{30}=\frac{23}{30}$
Probability that the marks of the chosen student are 31 or less $=\frac{4+3}{30}=\frac{7}{30}$
Probability that the marks of the chosen student lie in the interval 21 - 30 $=\frac{6}{30}=\frac{1}{5}$

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Question 53 Marks

In a survey of $200$ ladies, it was found that $142$ like coffee, while $58$ dislike it.
Find the probability that a lady chosen at random:
$i.$Likes coffee $ii.$Dislikes coffee.

Answer

Total number of tosses $= 200$ Number of ladies who likes coffee $= 142$ Number of ladies who dislikes coffee $= 58$ Let $E_1$ and $E_2$ be the events that the selected lady likes and dislikes coffee, respectively. Then,
$i.\ P($selected lady likes coffee$) = P(E_1) =\frac{\text{Number of ladies who like coffee}}{\text{Total number of laies}}$
$=\frac{142}{200}=0.71$
$ii.\ P($selected lady dislikes coffee $) = P(E_1) =\frac{\text{Number of ladies who dislike coffee}}{\text{Total number of laies}}$
$=\frac{58}{200}=0.29$
Remark: In the given survey, the only possible outcomes are $E_1$ and $E_2$ and $P(E_1) + P(E_2) = (0.71 + 0.29) = 1.$

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Question 63 Marks

A coin is tossed 500 times and we get:
heads: 285 times and tails: 215 times.
When a coin is tossed at random, what is the probability of getting

A head?
A tail?

Answer

Total number of tosses = 500 Number of heads = 285 Number of tails = 215

Let E be the event of getting a head.

P(getting a head) = $\text{P(E)}=\frac{\text{Number of heads coming up}}{\text{Total number of trials}}$
$=\frac{285}{500}=0.57$

Let F be the event of getting a tail.

P(getting a tail) = $\text{P(E)}=\frac{\text{Number of heads coming up}}{\text{Total number of trials}}$
$=\frac{215}{500}=0.43$

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Question 73 Marks

It is known that a box of 800 electric bulbs contains 36 defective bulbs.One bulb is taken at random out of the box. What is the probability that the bulb chosen is non-defective?

Answer

Total number of electric bulbs = 800
Number of defective bulbs = 36
⇒ Number of non-defective bulbs = 800 - 36 = 764
Hence,
Probability that the bulb chosen is non-defective $=\frac{\text{Number of non-defective bulbs}}{\text{Total number of bulbs}}$
$=\frac{764}{800}$
$=\frac{191}{200}$

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