Test 15 || Class 10th Maths || Ch.15 Probability1 (Match Special)

Test 15 || Class 10th Maths || Ch.15 Probability1 (Match Special)

इन प्रश्नों को हल करने से पहले हम कुछ बातें आपको बताना चाहेंगे।

सैंपल स्पेस (sample space) के लिए हम कैपिटल एस {S} यूज करेंगे। 

नंबर ऑफ सैंपल स्पेस के लिए हम एन एस   { n(S) } यूज करेंगे। 

फेवरेबल आउटकम के लिए हम ई {E} यूज करेंगे। 

नंबर ऑफ फेवरेबल आउट करने के लिए हम एन ई  { n(E) } यूज करेंगे तथा प्रोबेबिलिटी के लिए हम पी ई  { P(E) } यूज करेंगे।

Symbol used in Probability 

* Sample Space = S

* Number of Samples Space = n(S)

* Favourable Outcome = E

* Number of Favourable Outcomes = n(E)

* Probability = P(E)

Q.1 In a family of two children find the probability of having at least one girl.

[Board Term-2, 2012]

Sol. All possible outcomes,

S = {GG, GB, BG, BB}

Total number of possible outcomes,

n(S) = 4

Thus having at least one girl

Favourable outcome from sample Favourable outcomes are GG, GB and BG.

n(E) = 3

P(at least one girl),

P(E)  = Fevrable out came n(E)/ total out game n(S)

P(E)= 3/4

Q.2 In a family of three children find the probability of 

(i) having at least one boy

(ii) having at least two boys 

(iii) having at least three boys 

(iv) having at least one girl 

(v) having at least two girls 

(vi) having at least three girls

Sol. 

let the Boys = b  

and 

Girls = g

sample space:

bbb, bbg, bgg, ggg, gbb, ggb, gbg, bgb.

(i) having at least one boy

Favourable outcome from sample space:

bbb, bbg, bgg, gbb, ggb, gbg, bgb

P(E)  = Fevrable out came / total out game

P(E)= 7/8

(ii) having at least two boys

Favourable outcome from sample space: bbb, bbg, gbb, bgb

P(E)  = Fevrable out came / total out game

P(E)= 4/8 

=1/2

(iii) having at least three boys

Favourable outcome from sample space: bbb

P(E)  = Fevrable out came / total out game

P(E)= 1/8 

(iv) having at least one girl

Favourable outcome from sample space:

bbg, bgg, gbb, ggg, ggb, gbg, bgb

P(E)  = Fevrable out came / total out game

P(E)= 7/8

(v) having at least two girls

Favourable outcome from sample space:  bgg, ggg, ggb, gbg

P(E)  = Fevrable out came / total out game

P(E)= 4/8 

=1/2

(vi) having at least three girls

Favourable outcome from sample space:  ggg

P(E)  = Fevrable out came / total out game

P(E)= 1/8 

Q.3 A box contains 12 balls of which some are red in colour. If 6 more red balls are put in the box and a ball is drawn at random the probability of drawing a red ball doubles then what it was before. Find the number of the red balls in the box. 

Sol.

let the red balls in box = x

total ball in box = 12

P(E)1 = x/12

after adding 6 more bolls

total bones = 12 + x

P(E)2= (12+x)/18

ATQ

P(E)2=2×P(E)1

(12+x)/18=2×(x/12)

x=3

Q.4 A number x is selected from the numbers 1, 2, 3 and then a second number y is randomly selected from the numbers 1, 4, 9 then what is the probability that the product xy of the two numbers will be less than 9.

Sol.

Number x can be selected in three ways and corresponding to each such way there are three ways 

of selecting number y .

Therefore two numbers can be selected in 9 ways as 

(1, 1), (1, 4), (1, 9), 

(2, 1), (2, 4), (2, 9), 

(3, 1), (3, 4), (3, 9) 

So, total numbers of possible outcomes are 9.

The product xy will be less than 9, if x and y are chosen in one of the following ways: 

(1, 1), (1, 4), (2, 1), (2, 4), (3, 1)

P(E) = favourable outcome / total outcome

P(E) = 5/9

Q.5 Number x is selected at random from the numbers 1, 2, 3 and 4 another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that the product of x and y is less than 16.

Sol. 

x = 1, 2, 3, 4

y = 1, 4, 9, 16

After pairings

(1, 1)  (1, 4)  (1, 9)  (1, 16)

(2, 1)  (2, 4)  (2, 9)  (2, 16)

(3, 1)  (3, 4)  (3, 9)  (3, 16)

(4, 1)  (4, 4)  (4, 9)  (4, 16)

Product of x and y is less than zero

(1, 1) (1, 4) (1, 9) (2, 1) (2, 4) (3, 1) 

(3, 4) (4, 1) 

Probability that the product of x and y is less than 16.

P(E) = favourable outcome / total outcome

P(E) = 8/16

P(E) = 1/2

Q.6 If the probability of an event is p, then the probability of its complementary event will be

(a) p – 1 (b) p  (c) 1 – p (d) 1–1/p

Sol. 

P(E) = p

Since, 

P(E ) + P (E' ) = 1

P (E' ) = 1 – P(E )

P (E' ) = 1 – p

Thus (c) is correct option.

Q.7 Someone is asked to take a number from 1 to 100. The probability that it is a prime, is

(a) 8/25  (b) 1/4  (c) 3/4  (d) 14/50

Sol. 

Prime numbers between 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97, i.e 25 outcome.

n(S) = 100

n(E) = 25

P(E) = n(E)/n(S)

P(E) = 25/100

= 1/4

Q.8 A single letter is selected at random from the word PROBABILITY. What is the probability that the selected letter is a vowel.

Sol.

There are 11 letter in word PROBABILITY. Out of these 11 letter, 4 letter (O,A,I,I) are vowels.

n(S) = 11

n(E) = 4

Required probability,

P(E)= P(E)/P(S)=4/11

Q.9 If a number x is chosen a random from the number –3, –2, –1, 0, 1, 2, 3. What is probability that x² ≤ 4 ?

[Board 2020 Delhi Standard]

Sol.  

We have 7 possible outcome. 

Thus n(S) = 7

Favourable outcomes are 

–2, –1, 0, 1, 2 i.e. ≤ 5.

n(E) = 5

P(E)= n(E)/n(S)=5/7

Q.10 The probability of getting a bad egg in a lot of 400 is 0.035. what is the number of bad eggs in the lot.

Sol. 

We have n(S) = 400

P(E)= 0.035

n(E) = ?

P(E)= n(E)/n(S)

n(E) = P(E) × n(S)

n(E) = 0.035 x 400 

n(E) = 14

Q.11 What is the probability that a non-leap year has 53 Mondays ?

[Board Term-2, 2015]

Sol. 

There are 365 days in a non-leap year.

365 days = 52 weeks + 1 day

One day can be 

M, T, W, Th, F, S, S i.e. total 7 

Possible outcomes and only one favourable outcome.

Thus n(S) = 7 and n(E) = 1

P(53 Mondays in non-leap year)

P(E)= n(E)/n(S)=1/7

Q.12 Find the probability that a leap year has 53 Sundays.

[Board Term-2, 2012]

Sol. 

366 days = 52 weeks + 2 days

2 days can be 

MT, TW, WTh, ThF, FS, SS, SM out 

of which SS and SM are favourable outcome.

Total number of possible outcomes,

n(S) = 7

Thus number of favourable outcome,

n(E) = 2

P(53 Mondays in a leap year)

P(E)= n(E)/n(S)=2/7

Q.13 Find the probability that 5 Sundays occur in the month of November of a randomly selected year.

[Board 2020 Delhi Basic]

Sol. 

Total no. of days in November = 30

So, it has 4 weeks and 2 days. 

4 weeks have 4 Sundays. 

The two remaining days should be

1. Sunday, Monday

2. Monday, Tuesday

3. Tuesday, Wednesday

4. Wednesday, Thursday

5. Thursday, Friday

6. Friday, Saturday

7. Saturday, Sunday

Thus number of possible outcomes,

n(S) = 7

Number of favourable outcome,

n(E) = 2

So, the probability of getting 5 Sunday in the month of November,

P(E)= n(E)/n(S)=2/7

Q.14 From the number 3, 5, 5, 7, 7, 7, 9, 9, 9, 9, one number is selected at random, what is the probability that the selected number is mean?

[Board Term-2, 2012]

Sol.

Total outcomes, n(S)= 10

Mean = (3+5+5+7+7+7+9+9+9+9)/10

= 70/10 

= 7 

Thus 7 is the mean of given numbers and frequency of 7 is 3 in given data.

Number of favourable outcomes,

n(E) = 3

P(mean), 

P(E)= n(E)/n(S)=3/7

Q.15 A letter of English alphabet is chosen at random, find the probability that the letter so chosen is :

(i) a vowel,

(ii) a consonant.

[Board Term-2 Delhi 2014]

Sol. 

Since total number in English alphabet is 26, in which 5 vowels and 21 consonants. 

In both case total possible outcome

n(S) = 26

(i) a vowel, n(E) = 5

P(E) for a vowel =n(E)/n(S)

P(E)= 5/26

(ii) a consonant, n(S) = 26

P(E) for a consonant =n(E)/n(S)

P(E)= 21/26