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given: pack of 52 playing cards

Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\) 

five cards are drawn at random, therefore 

total possible outcomes are ⁵²C₅ 

therefore n(S)=2598960 

(i) let E be the event that exactly only one ace is present 

n(E)= ⁴C₁ ⁴⁸C₄ = 778320

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{778320}{2598960}\) = \(\frac{3243}{19829}\) 

(ii) let E be the event that at least one ace is present 

E= {1 or 2 or 3 or 4 ace(s)} 

n(E)= ⁴C₁ ⁴⁸C₄ + ⁴C₂ ⁴⁸C₃ + ⁴C₃ ⁴⁸C₂ + ⁴C₄ ⁴⁸C₁ = 886656

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{886656}{2598960}\) = \(\frac{18472}{54145}\) 

Given: Five cards are drawn from a pack of 52 cards.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Five cards are drawn at random,

Total possible outcomes are ⁵²C₅

n (S) = 2598960

(i) Let E be the event that exactly only one ace is present

n (E) = ⁴C₁⁴⁸C₄ = 778320

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

= 778320 / 2598960

= 3243/10829

(ii) Let E be the event that at least one ace is present

E = {1 or 2 or 3 or 4 ace(s)}

n (E) = ⁴C₁⁴⁸C₄ + ⁴C₂⁴⁸C₃ + ⁴C₃⁴⁸C₂ + ⁴C₄⁴⁸C₁ = 886656

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

= 886656 / 2598960

= 18472/54145

Given: A box contains 100 bulbs, 20 of which are defective.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Ten bulbs are drawn at random for inspection,

Total possible outcomes are ¹⁰⁰C₁₀

n (S) = ¹⁰⁰C₁₀

(i) Let E be the event that all ten bulbs are defective

n (E) = ²⁰C₁₀

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

= ²⁰C₁₀ / ¹⁰⁰C₁₀

(ii) Let E be the event that all ten good bulbs are selected

n (E) = ⁸⁰C₁₀

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

= ⁸⁰C₁₀ / ¹⁰⁰C₁₀

(iii) Let E be the event that at least one bulb is defective

E= {1,2,3,4,5,6,7,8,9,10} where 1,2,3,4,5,6,7,8,9,10 are the number of defective bulbs

Let E′ be the event that none of the bulb is defective

n (E′) = ⁸⁰C₁₀

P (E′) = n (E′) / n (S)

= ⁸⁰C₁₀ / ¹⁰⁰C₁₀

So, P (E) = 1 – P (E′)

= 1 – ⁸⁰C₁₀ / ¹⁰⁰C₁₀

(iv) Let E be the event that none of the selected bulb is defective

n (E) = ⁸⁰C₁₀

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

= ⁸⁰C₁₀ / ¹⁰⁰C₁₀

Given: There are four men and six women on the city councils.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

From the city council one person is selected as a council member so, we have to find the probability that it is a woman.

Total possible outcomes of selecting a person is ¹⁰C₁

n (S)= ¹⁰C₁ = 10

Let E be the event that it is a woman

n (E) = ⁶C₁ = 6

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

= 6 / 10

= 3/5

Answer is (B) 1/3

Sample space of children in a family is :

S = {B, G, G, B, B, G, B, B, G}

Event of 2 boys and 1 girl A = {B, B, G}

∴ n(S) = 3, n(A) = 1

Thus, required probability = n(A)/n(S) = 1/3

Answer: (c)  \(\frac{10}{19}\)

Let the events E_1, E_2, and A be defined as follows: 

E₁ = Choosing bag A 

E₂ = Choosing bag B 

A = Choosing red ball. 

Then, P(E₁) = P(E₂) = \(\frac{1}{2}\) 

(∵ There are two bags that have an equally likely chance of being chosen) 

P(A/E₁) = P(Drawing red ball from bag A) = \(\frac{2}{4} =\frac{1}{2}\)  (∵ 2 red out of 4 balls) 

P(A/E₂) = P(Drawing red ball from bag B) = \(\frac{5}{9}\)        (5 red out of 9 balls)

∴ P(Red balls are drawn from bag B)

\(=P(E_2/A) = \frac{P(E_2)\times P(A/E_2)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}\)

\(=\frac{\frac{1}{2}\times \frac{5}{9}}{\frac{1}{2}\times \frac{1}{2}+ \frac{1}{2}\times \frac{5}{9}}\) = \(\frac{\frac{5}{18}}{\frac{1}{4}+\frac{5}{18}}\) = \(=\frac{5/18}{19/36}\) = \(\frac{10}{19}\)

 Answer: (a) \(\frac{1}{13}\)

Let S be the sample space of drawing a card out of 52 cards. 

Then, n(S) = 52

Let A: Event of drawing a king ⇒ n(A) = 4                                (A pack has 4 kings) 

B: Event of drawing a black card ⇒ n(B) = 26                           (A pack has 26 black cards) 

A ∩ B: Event of drawing a king of a black card 

⇒ n(A ∩ B) = 2                                                                            (A pack has 2 black kings)

\(\therefore P(A) =\frac{4}{52}= \frac{1}{13}\) ,\( P(B) =\frac{26}{52}= \frac{1}{2}\)  , \(P(A\,\cap\,B) = \frac{2}{52} = \frac{1}{26}\) 

P(Getting a king, given card is drawn is black

\(P(A/B) =\frac{P(A\,\cap\,B)}{P(A)} =\frac{1/26}{1/2} =\frac{2}{26}\) 

\(\frac{1}{13}\)    (∵ Event A depends on B)

Answer is (C) 1

Answer is (B) 1/2

Required probability

= ²⁶C₁/⁵²C₁ = 26/52

= 1/2

n(E) = Total ways to get a queen x Total ways to get a queen x Total ways to get a fack x Total ways to get 1 ace

= 4 × 4 × 4 × 4 = 256.

Sample space obtained by throwing a coin four time

S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTP, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}

n(S) = 16

Let appearance of tail at least 3 times is event A, then

A = {HHHH, HHHT, HHTH, HTHT, HTHH}

n(A) = 5

Thus, required probability

P(A) = n(A)/n(S) = 5/16

Thus, probability to get tail at least 3 times is 5/16

Answer is (C) 4

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, ITT)

n(E) = 4

After drawing first bulb two results D of will be

After drawing 2nd bulb result

= (D,N) × (D,N)

= {DD, DN, ND, NN}

After drawing 3rd bulb result

= (D, N) × {DD, ND, NN}

= {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}

Thus sample space

S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}

Answer is (C) Sample point

Answer is (A) 12

No. of elements = n{H, T} × n{1, 2,3,4,5,6}

= 2 × 6 = 12

Given that p is the prob. that coin shows a head then 1 – p will be the prob. that coin shows a tail.

Now α = P (A gets the 1^st head in 1^st try)

⇒ α = P(H) + P(T) P(T) P(T) P(H) + P (T) P (T) P (T) P (T) P (T) P (T) P(H)

= p + (1 – p)³p + (1 – p)⁶p+ . . . . . . . . . . . . . . . . . . . . . 

= p[1 + (1 – p)³ (1 – p)⁶ + . . . . . . . . . . . . . . . . . . . . . . . ]

= p/1 – (1 – p)³ 

NOTE THIS STEP . . . (i)

Similarly β = P (B gets the 1^st head in 1^st try) + P (B gets the 1^st head in 2^nd try) + . . . . . . . . . . . . . . . . . . 

= P(T) P(H) + p(T) p(T) p(T) p(T) p(T) P(H) + . . . . . . . . . . . . . . . . . . .

= (1 – p) p + (1 – p)⁴ + . . . . . . . . . . . . . . . . . . . . 

= (1 – p) p/1 – (1 – p)³ . . . . . . . . . . . . . . . . . . . . . (ii) 

From (i) and (ii) we get β = (1 – p)α

Also (i) and (ii) give expression for α and β in terms of p.

Also α + β + γ = 1(exhaustive events and mutually exclusive events)

⇒ γ = 1 – α – β = 1 - α – (1 – p) α

= 1 – (2 – p)α

= 1 – (2 – p)p/1 – (1 - p)³ 

= 1 – (1 – p)³ – (2p – p²)/1 – (1 – p)³

= 1 – 1 + p³ + 3p(1 – p) – 2p + p²/1 – (1 – p)³

= p³ – 2p² + p/1 – (1 – p)³

= p(p² – 2p + 1)/1 – (1 – p)³

= p (1 – p) 2/1 – (1 – p)³

(i) Given : P(A) = 0.25, P(A or B) = 0.5 and P(B) = 0.4 

To find : P(A and B) 

Formula used : P(A or B) = P(A) + P(B) - P(A and B) 

Substituting in the above formula we get, 

0.5 = 0.25 + 0.4 – P(A and B) 

0.5 = 0.65 - P(A and B) 

P(A and B) = 0.65 – 0.5 

P(A and B) = 0.15 

P(A and B) = 0.15 

(ii) Given : P(A) = 0.25, P(A and B) = 0.15 ( from part (i)) 

To find : P(A and \(\overline{B}\) ) 

Formula used : P(A and \(\overline{B}\) ) = P(A) – P(A and B) 

Substituting in the above formula we get,

P(A and \(\overline{B}\) ) = 0.25 – 0.15 

P(A and \(\overline{B}\) ) = 0.10 

P(A and \(\overline{B}\)) = 0.10

Given : \(P( \bar{A})\) = 0.4, P(A or B) = 0.7 and P(A and B) = 0.3 

To find : P(B) 

Formula used : P(A) = 1 – \(P( \bar{A})\) 

P(A or B) = P(A) + P(B) - P(A and B) 

We have \(P( \bar{A})\) = 0.4 

P(A) = 1 – 0.4 = 0.6 

We get P(A) = 0.6 

Substituting in the above formula we get, 

0.7 = 0.6 + P(B) – 0.3 

0.7 = 0.3 + P(B) 

0.7 – 0.3 = P(B) 

0.4 = P(B) 

P(B) = 0.4

P(A) = 1/2, P(B) = 1/3

We know, P(A or B) = P(A) + P(B) – P(A and B)

For mutually exclusive events A and B, P(A and B) = 0

P(A or B) = 1/2 + 1/3 – 0

P(A or B) = 5/6

Let, event A: The first ball drawn is white 

event B: Second ball drawn is white. 

P(A) = 5/8

After drawing the first ball, without replacing it into the bag a second ball is drawn from the remaining 7 balls.  

∴ P(B/A) = 4/7

∴ P(Both balls are white) = P(A ∩ B) 

= P(A) . P(B/A)

= 5/8 x 4/7

= 5/14

B. 2/7

Explanation:

We know that in a non-leap year, there are 365 days and we know that there are 7 days in a week

∴ 365 ÷ 7 = 52 weeks + 1 day

This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday

∴ Total Outcomes = 7

If this day is a Tuesday or Wednesday, then the year will have 53 Tuesday or 53 Wednesday.

∴P (non-leap year has 53 Tuesdays or 53 Wednesdays) = 1/7 + 1/7 = 2/7

Hence, the correct option is (B).

A leap year has 366 days (52 weeks + 2 days) 

That two days may be (sun, mon) (mon, tue) (tue, wed) (wed, thurs) (thurs, fri) (fri, sat) (sat, sun) 

∴ n(S) = 7 

Number of events for the occurrence of 53 Tuesdays and 53 Mondays, n(E) = 1

∴ \(\frac{n(E)}{n(S)}\) = \(\frac{1}{7}\)

A non-leap year consists of 365 days. Therefore in a non-leap year there are 52 complete weeks and 1 day over which can be one of the seven days of the week. 

Possible outcomes n(S) = 7 = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}. 

∴ Number of possible outcomes n(S) = 7 

(As there are seven days in a week) 

Let A : Getting the extra day as Sunday or Tuesday or Thursday 

⇒ A = {Sunday, Tuesday, Thursday} 

⇒ n(A) = 3

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{3}{7}.\)

In a single throw of die any one of six numbers 1,2,3,4,5,6 can be obtained. Therefore, the tome number of elementary events associated with the random experiment of throwing a die is 6.

(i) Let A denote the event “Getting a prime no”. Clearly, event A occurs if any one of 2,3,5 comes as out come.

So, Favorable number of elementary events = 3

Hence, P (Getting a prime no.) =3/6=1/2

(ii) An multiple of 2 or 3 is obtained if we obtain one of the numbers 2,3,4,6 as out comes

So, Favorable number of elementary events = 4

Hence, P (Getting multiple of 2 or 3) =4/6=2/3

(iii) The event “Getting a number greater than 3” will occur, if we obtain one of number 4,5,6 as an out come.

So, Favorable number of out comes = 3

Hence, required probability =3/6=1/2

(i) When one coin is tossed twice, the sample space is 

S = {HH,HT,TH,TT}. 

Let X denotes, the number of heads in any outcome in S, 

X (HH) = 2, X (HT) = 1, X (TH) = 1 and X (TT) = 0 

Therefore, X can take the value of 0, 1 or 2. It is known that 

P(HH) = P(HT) = P(TH) = P(TT) = 1/4 

∴ P(X = 0) = P (tail occurs on both tosses) = P({TT}) = 1/4 

P(X = 1) = P (one head and one tail occurs) = P({TH,HT}) = 2/4 = 1/2 

and P(X = 2) = P (head occurs on both tosses) = P({HH}) = 1/4 

Thus, the required probability distribution is as follows

(ii) When three coins are tossed thrice, the sample space is 

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} which contains eight equally likely sample points. 

Let X represent the number of tails. Then, X can take values 0, 1, 2 and 3. P(X = 0) = P (no tail) = P({HHH}) = 1/8 , P (X = 1) = 

P (one tail and two heads show up) = P({HHT,HTH,THH}) = 3/8,

P (X = 2) = P (two tails and one head show up) = P({HTT,THT,TTH}) = 3/8 

and P(X = 3) = P (three tails show up) = P({TTT}) = 1/8 

Thus, the probability distribution is as follows

When two coins are tossed simultaneously, the possible outcomes are(H, H),(H, T),(T, H),(T, T) which are all equally likely. Here (H, H) means head up on the first coin (say on Rs.1) and head up on the second coin (Rs.2).Similarly (H, T) means head up on the first coin and tail up on the second coin and so on.

The outcomes favorable to the event E,at least one head' are (H, H), (H, T) and (T, H).

So. the number of outcomes favorable to.E is 3.

Therefore,  P(E)=3/4

i.e., the probability that Harpreet gets at least one head is 3/4.

Possible outcomes as HH, TT, TH, HT

(i) P(E₁) = 3/4

(ii) P(E₂) = 2/4 = 1/2

We write H for ‘head’ and T for ‘tail’. When two coins are tossed simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all equally likely. Here (H, H) means head up on the first coin (say on Re 1) and head up on the second coin (Rs 2). Similarly (H, T) means head up on the first coin and tail up on the second coin and so on.

The outcomes favourable to the event E, ‘at least one head’ are (H, H), (H, T) and (T, H).

So, the number of outcomes favourable to E is 3.

Therefore,

P(E)=3/4

i.e., the probability that Harpreet gets at least one head is 3/4.

Total number of possible outcomes, n(S) = 4 

Number of favorable outcomes, 

n(E) = 3

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{3}{4}\)

Total number of possible outcomes, n(S) = 5 

(i) Number of favorable outcomes, 

n(E) = 1

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{1}{5}\)

(ii) If queen is drawn and put aside, total number of remaining cards, n(S) = 4 

a) Number of favorable outcomes, 

n(E) = 1

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{1}{4}\)

b) Total number of events of getting a queen = 0 as there is no queen

∴ P(E) = 0

Total no. of possible outcomes when 2 dice are thrown = 6×6 = 36 which are

{ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }

(i) E ⟶ event of getting sum that turn up is 8

No. of possible outcomes = 36

No. of favourable outcomes = 5 {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}

P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 5/36

(ii) Let E ⟶ event of obtaining a total of 6

No. of favourable outcomes = 5

{(1, 5) (2, 4) (3, 3) (4, 2) (5, 1)}

P(E) = 5/36

(iii) Let E ⟶ event of obtaining a total of 10.

No. of favourable outcomes = 3 {(4, 6) (5, 5) (6, 4)}

P(E) = 3/36 = 1/12

(iv) Let E ⟶ event of obtaining the same no. on both dice

No. of favourable outcomes = 6 {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}

P(E) = 3/36=1/12

(v) E ⟶ event of obtaining a total more than 9

No. of favourable outcomes = 6 {(4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6)}

P(E) = 6/36 = 1/6

(vi) The maximum sum is 12 (6 on 1^st + 6 on 2^nd)

So, getting a sum of no’s appearing on the top of the two dice as 13 is an impossible event.

∴ Probability is 0

(vii) Since, the sum of the no’s appearing on top of 2 dice is always less than or equal to 12, it is a sure event.

Probability of sure event is 1.

So, the required probability is 1.

The correct option is: (B)

Explanation:

 Outcomes are 13, 14,15,......., 60.

 Total number of possible outcomes = 60 - 12 = 48 

(i) The numbers divisible by 5 are 15,20,25,30, 35, 40, 45, 50, 55, 60.

Thus, the number of numbers divisible by 5 = 10 

Required Probability = 10/48 = 5/24

(ii) Perfect square numbers are 16,25,36,49 

Thus, the number of perfect square number = 4

Required probability = 4/48 = 1/12

Total no. of possible outcomes = 5 {5 cards}

(i) E ⟶ event of getting a good pen.

No. of favourable outcomes = 132 {132 good pens}

P (E) = (No.of favorable outcomes)/(Total no.of possible outcomes)

∴ P(E) = 1/5

(ii) If queen is drawn & put aside,

Total no. of remaining cards = 4

(a) E ⟶ event of getting a queen.

No. of favourable outcomes = 1 {1 ace card}

Total no. of possible outcomes = 4 {4 remaining cards}

P(E) = 1/4

(b) E ⟶ event of getting a good pen.

No. of favourable outcomes = 0 {there is no queen}

P(E) = 0/4=0

∵ E is known as impossible event.

A non-leap year has 365 days

A year has 52 weeks. Hence there will be 52 Sundays for sure.

52 weeks = 52 x 7 = 364 days .

365– 364 = 1day extra.

Remaining 1 days may be any day from Monday to Sunday = 1/7

Thus probability of being Sunday

= 1 – 1/7 = 6/7

Here we should take probability of not being Sunday here we should take probability of not being Sunday

∴ Required Probability = 6/7

The correct option is:

The correct option: (B) 5/108

Explanation:

Total number of outcomes when three dice are thrown = 6 x 6 x 6 = 216 

For sum of numbers to be 15, 

possible ways are, (6, 6, 3), (6, 3,6), (3, 6,6), (6, 5,4), (6,4, 5), (5,4, 6), (5, 6,4), (4, 5, 6), (4, 6, 5), (5, 5, 5) 

.'. Number of favorable outcomes = 10 

.'.  Required probability = 10/216 = 5/108

The non-leap year has 365 days in it. 

There are 52 weeks in a year 

i.e., 52 Sundays.

The remaining 1 day may be either, Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.

∴ Total number of possible outcomes = 7

The number of net falling Sunday = 6

∴ The probability of falling only 52 Sundays in a non-leap year = \(\frac { 6 }{ 7 }\)

The correct option: (A) 2/5

Explanation:

Total number of marbles = 10

.'. Probability of drawing a green marble = 4/10 = 2/5

The correct option is: (D) 7/9

Explanation:

Total number of marbles = 3 + 4 + 2 = 9

No. of green and blue marbles = 3 + 4 = 7.

.'. Probability of not getting an orange marble = 7/9

A month of 30 days has four complete weeks and 2 extra days. These two days may be;

(i) Monday and Tuesday

(ii) Tuesday and Wednesday

(iii) Wednesday and Thursday

(iv) Thursday and Friday

(v) Friday and Saturday

(vi) Saturday and Sunday

(vii) Sunday and Monday.

∴ The number of all possible events = 7

Now in these events, last two events have Sunday each.

∴ Number of favourable outcomes = 2

∴ In these remaining two days, the probability of having Sunday = \(\frac { 2 }{ 7 }\).