Probability Test

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Probability Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

When a coin is tossed at random, then the probability of getting a head is ________.

A
$0$

B
$\dfrac {1}{2}$

C
$1$

D
$2$

Solution

Tossing of a coin at random can give only two results, i.e. Heads or Tails.
$\Rightarrow$ Total number of outcomes possible $= 2$

Probability $= \dfrac{\text{Favourable Outcomes}}{\text{Total Outcomes}} = \dfrac{1}{2}$
Question 2
1 / -0

Two men hit at a target with probabilities $\dfrac{1}{2}$ and $\dfrac{1}{3}$ respectively. What is the probability that exactly one of them hits the target?

A
$\dfrac{1}{2}$

B
$\dfrac{1}{3}$

C
$\dfrac{1}{6}$

D
$\dfrac{2}{3}$

Solution

${ E }_{ 1 }$ denotes the event when first man hits the target
${ E }_{ 2}$ denotes the event when second man hits the target
$P({ E }_{ 1 })=\cfrac { 1 }{ 2 } ,P({ E }_{ 2 })=\cfrac { 1 }{ 3 }$
Probability that exactly one of them hits the target, $P=P({ E }_{ 1 })P({ \overset { \_ }{ E } }_{ 2 })+P({ E }_{ 2 })P({ \overset { \_ }{ E } }_{ 1 })$
$=\dfrac { 1 }{ 2 } \times \dfrac { 2 }{ 3 } +\dfrac { 1 }{ 3 } \times \dfrac { 1 }{ 2 } =\dfrac { 1 }{ 2 }$
Question 3
1 / -0

In a single throw of a die, the probability of getting a multiple of $3$ is ____________.

A
$\displaystyle\frac{1}{2}$

B
$\displaystyle\frac{1}{3}$

C
$\displaystyle\frac{1}{6}$

D
$\displaystyle\frac{3}{4}$

Solution

Possible outcomes on rolling a dice are $1,2,3,4,5,6$
Out of the possible outcomes, multiple of $3$ are $3$ and $6$ .

Probability $=$ Favourable Outcomes $\div$ Total Outcomes $= \dfrac{2}{6} = \dfrac{1}{3}$
Question 4
1 / -0

A die is tossed $80$ times and the number $3$ is obtained $14$ times. Now, a dice is tossed at random, then the probability of getting the number $3$ is ________.

A
$\dfrac {7}{40}$

B
$\dfrac {3}{14}$

C
$\dfrac {1}{2}$

D
$\dfrac {3}{40}$

Solution

Total number of times dice was rolled $= 80$
Number of times $3$ is obtained $= 14$

Probability $= \dfrac{\text{Favourable Outcomes}}{\text{Total Outcomes}}$

$\Rightarrow$ Probability of obtaining $3 = \dfrac{14}{80} =\dfrac{7}{40}$
Question 5
1 / -0

A die is thrown $400$ times, the frequency of the outcomes of the events are given as under.
outcome
$1$
$2$
$3$
$4$
$5$
$6$
Frequency
$70$
$65$
$60$
$75$
$63$
$67$
Find the probability of occurrence of an odd number.

A
The probability of occurrence of odd number $=\dfrac{5}{7}$

B
The probability of occurrence of odd number $=\dfrac{9}{2}$

C
The probability of occurrence of odd number $=\dfrac{193}{400}$

D
The probability of occurrence of odd number $=\dfrac{200}{299}$

Solution

Sum of frequencies $= 400$

Odd numbers are $1, 3, 5$

Therefore, frequency of all odd numbers $= 70 + 60 + 63 = 193$

$P(\text{event})=\dfrac{\text{Frequency of occurring of event}}{\text{The total number of trials}}$

Therefore , probability of occurrence of odd number $=\dfrac{193}{400}$
Question 6
1 / -0

The probability that a two digit number selected at random will be a multiple of ' $3$ ' and not a multiple of ' $5$ ' is

A
$\displaystyle \frac{2}{15}$

B
$\displaystyle \frac{4}{15}$

C
$\displaystyle \frac{1}{15}$

D
$\displaystyle \frac{4}{90}$

Solution

$\Rightarrow$ Total number of two digit number $=90$

$\Rightarrow$ Total number of two digit number which is divisible by $3=30$

Out of this there are $6$ numbers divisible by $15$ $(15,30,45,60,75,90)$ which are also divisible by $5.$

$\Rightarrow$ Total two digit number which are divisible by $3$ but not $5=30-6=24$

$\Rightarrow$ Required probability $=\dfrac{24}{90}=\dfrac{4}{15}$
Question 7
1 / -0

On one page of a telephone directory there were $200$ telephone numbers. The frequency distribution of their unit place digit is given in the following table:
Digit $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$
Frequency $22$ $26$ $22$ $22$ $20$ $10$ $14$ $28$ $16$ $20$

A

$0.15$

B

$0.17$

C

$0.18$

D
None of these

Solution

The probability that the digit in its unit place is more than $7 = \dfrac {16+20}{200} =\dfrac {36}{200} = 0.18$
Question 8
1 / -0

In a survey of $364$ children aged $19-36$ months, it was found that $91$ liked to eat potato chips. If a child is selected at random, the probability that he/she does not like to eat potato chips is :

A
$0.25$

B
$0.50$

C
$0.75$

D
$0.80$

Solution

Favorable cases $=364 - 91 =273$
Total cases $= 364$
$\therefore$ Probability $= \dfrac { 273 }{ 364 }=0.75$
Question 9
1 / -0

A coin is tossed $150$ times and the outcomes are recorded. The frequency distribution of the outcomes $H$ (i.e., head) and $T$ (i.e., tail) is given below :
Outcome $H$ $T$
Frequency $85$ $65$
Find the value of $P(H)$ , i.e., probability of getting a head in a single trial.

A
$P(H) =0.769$ (approx)

B
$P(H) =0.663$ (approx)

C
$P(H) =0.567$ (approx)

D
None of these

Solution

Total number of trials $= 150$
Chances or trials which favour the outcome $H=85$
$P(H) =$ $\displaystyle \frac{85}{150}$ $=$ $0.567$ (approx)
Question 10
1 / -0

To know the opinion of the student about the subject statistic, a survey of $200$ students was conducted.
The data is recorded in the following table
Opinion Like Dislike
No. of Students $135$ $65$
Find the probability that a student chosen at random
$(i)$ likes statistics, $(ii)$ does not like it.

A
$(i) \displaystyle \frac{13}{40}$

$(ii) \displaystyle \frac{19}{40}$

B
$(i) \displaystyle \frac{27}{40}$

$(ii) \displaystyle \frac{13}{40}$

C
$(i)\displaystyle \frac{17}{40}$

$(ii) \displaystyle \frac{29}{40}$

D
None of these

Solution

Total number of students $= 200$
(i) Number of students who like the subject of statistics $= 135$
The probability that a student likes that subject $=$ $\displaystyle\frac{135}{200}$ $=$ $\displaystyle \frac{27}{40}$

(ii) Number of students who dislike the subject of statics $=$ 65
The probability that a student dislikes the subject $=$ $\displaystyle \frac{65}{200}$ = $\displaystyle \frac{13}{40}$

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Re-Attempt Weekly Quiz Competition

Digit	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
Frequency	$22$	$26$	$22$	$22$	$20$	$10$	$14$	$28$	$16$	$20$

outcome	$1$	$2$	$3$	$4$	$5$	$6$
Frequency	$70$	$65$	$60$	$75$	$63$	$67$

Outcome	$H$	$T$
Frequency	$85$	$65$

Opinion	Like	Dislike
No. of Students	$135$	$65$

Probability Test - 16

Result

Probability Test - 16

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Question 1

000

12\dfrac {1}{2}21​

111

222

Solution

Question 2

12\dfrac{1}{2}21​

13\dfrac{1}{3}31​

16\dfrac{1}{6}61​

23\dfrac{2}{3}32​

Solution

Question 3

12\displaystyle\frac{1}{2}21​

13\displaystyle\frac{1}{3}31​

16\displaystyle\frac{1}{6}61​

34\displaystyle\frac{3}{4}43​

Solution

Question 4

740\dfrac {7}{40}407​

314\dfrac {3}{14}143​

12\dfrac {1}{2}21​

340\dfrac {3}{40}403​

Solution

Question 5

The probability of occurrence of odd number=57=\dfrac{5}{7}=75​

The probability of occurrence of odd number=92=\dfrac{9}{2}=29​

The probability of occurrence of odd number=193400=\dfrac{193}{400}=400193​

The probability of occurrence of odd number=200299=\dfrac{200}{299}=299200​

Solution

Question 6

215\displaystyle \frac{2}{15}152​

415\displaystyle \frac{4}{15}154​

115\displaystyle \frac{1}{15}151​

490\displaystyle \frac{4}{90}904​

Solution

Question 7

0.150.150.15

0.170.170.17

0.180.180.18

None of these

Solution

Question 8

0.250.250.25

0.500.500.50

0.750.750.75

0.800.800.80

Solution

Question 9

P(H)=0.769P(H) =0.769P(H)=0.769 (approx)

P(H)=0.663P(H) =0.663P(H)=0.663 (approx)

P(H)=0.567P(H) =0.567P(H)=0.567 (approx)

None of these

Solution

Question 10

(i)1340(i) \displaystyle \frac{13}{40} (i)4013​(ii)1940(ii) \displaystyle \frac{19}{40} (ii)4019​

(i)2740(i) \displaystyle \frac{27}{40} (i)4027​(ii)1340(ii) \displaystyle \frac{13}{40} (ii)4013​

(i)1740(i)\displaystyle \frac{17}{40} (i)4017​(ii)2940(ii) \displaystyle \frac{29}{40} (ii)4029​

None of these

Solution

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$0$

$\dfrac {1}{2}$

$1$

$2$

$\dfrac{1}{2}$

$\dfrac{1}{3}$

$\dfrac{1}{6}$

$\dfrac{2}{3}$

$\displaystyle\frac{1}{2}$

$\displaystyle\frac{1}{3}$

$\displaystyle\frac{1}{6}$

$\displaystyle\frac{3}{4}$

$\dfrac {7}{40}$

$\dfrac {3}{14}$

$\dfrac {1}{2}$

$\dfrac {3}{40}$

The probability of occurrence of odd number $=\dfrac{5}{7}$

The probability of occurrence of odd number $=\dfrac{9}{2}$

The probability of occurrence of odd number $=\dfrac{193}{400}$

The probability of occurrence of odd number $=\dfrac{200}{299}$

$\displaystyle \frac{2}{15}$

$\displaystyle \frac{4}{15}$

$\displaystyle \frac{1}{15}$

$\displaystyle \frac{4}{90}$

$0.15$

$0.17$

$0.18$

$0.25$

$0.50$

$0.75$

$0.80$

$P(H) =0.769$ (approx)

$P(H) =0.663$ (approx)

$P(H) =0.567$ (approx)

$(i) \displaystyle \frac{13}{40}$

$(ii) \displaystyle \frac{19}{40}$

$(i) \displaystyle \frac{27}{40}$

$(ii) \displaystyle \frac{13}{40}$

$(i)\displaystyle \frac{17}{40}$

$(ii) \displaystyle \frac{29}{40}$