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Correct option d (5/13)   

Explanation:

• Total Boy = 115 

• Total Girl = 65

• Probability that the student is a girl =P(G)= 65/180 

• Total Ex grade holders = 80 

• Probability that a student is a Ex grade holder = P(A) = 80/180

• The number of girls getting Ex grade = 25

• Probability that a student getting Ex grade and a girl = P(A∩G) = 25/180 

• Probability that a student getting Ex grade is a girl 

= P(A|G) =P (A ∩ G)/P(G) = 25/180/65/180 = 25/65 = 5/13

Correct option (b) 0.34

Explanation:

To Solve this problem use Additive law of Probability. 

P(A∪B)=P(A)+P(B)-P(A∩B)

P(C) = 0.25 

P(M) = 0.18 

P(C∩M) = 0.09 

P(at least one) = P(C or M) = P( C ∪ M) 

= 0.25 + 0.18 - 0.09 = 0.34

Correct option c (1-b, 2-c, 3-a)

Explanation:

Given:

Total number of balls in box = (4 blue + 5 black + 8 white) balls = 17 balls

Formula used:

Probability of event = (Number of favourable outcomes)/(Total number of outcomes)

Calculations:

Let,

E1 = event of picking a white ball

E2 = event of picking a blue ball

Then,

P(E1 ) = 8/17

P(E2) = 4/17

⇒  Probability of picking a cap that it is either blue or white = 4/17 + 8/17 = 12/17 

 the probability that it is neither white nor blue ball = 1 -12/17 = 5/17

∴ The probability that it isneither white nor blue ball is 5/17 

Given

Number of black balls = 7

Number of white balls = 6

Formula used

P(E) = Number of favourable outcomes/Total number of outcomes

Calculation

Probability of getting black ball = 7/13

Probability of getting white ball = 6/13

Required probability = (7/13) + (6/13)

⇒ 1

Clearly n(S)=6*6=36 

Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.Then 

E={(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2), (6,6)} 

n(E)=14. 

Hence p(e)=n(e)/n(s)=14/36=7/18

Given:

Die has = 6 numbers

Calculation:

Let O = Event of getting odd number

 ⇒ {1,3,5}

⇒ Let F = Event of getting a number greater than 2

⇒ {3,4,5,6}

⇒ O ᴒ F = {3,5}

⇒ n(O u F) = n(O) + n(F) – n(O ᴒ F)

⇒ 3 + 4 – 2

 ⇒ 5

∴ Required Probability = 5/6

Given:

Orange balls = 3

Yellow balls = 4

Purple balls = 7

Total balls = 14

Calculation:

Total number of balls is 14

⇒ Required Probability

⇒ (3 + 7)/14

⇒ 10/14

∴ required probability is 5/7

Let us denote 2 boys and 2 girls in room X as B1, B2 and G1, G2 respectively. Let us denote 1 boy and 3 girls in room Y as B3, and G3, G4, G5 respectively.
Accordingly, the required sample space is given by
S = {XB1, XB2, XG1, XG2, YB3, YG3, YG4, YG5}

Here S={1,2,3,4,5,6} 

Let E be the event of getting the multiple of 3 

then ,E={3,6} 

P(E)=n(E)/n(S)=2/6=1/3

Given:

A bag contains 2 red 3 green and 2 blue balls, Two balls are drawn at random. 

Formula Used:

Probability = favorable outcome/total outcome

Calculation:

None of the balls drawn is blue, this can only happen when the two balls drawn at random are either red and green or both.

Total number of balls = 2 + 3 + 2 = 7

⇒ Number of ways of drawing 2 balls out of 7 = ⁷C₂ = (7 × 6) / (2 × 1) = 42/2 = 21

Number of balls other than blue = 5

⇒ Number of ways of drawing 2 balls out of 5 = ⁵C₂ = (5 × 4) / (2 × 1) = 20/2 = 10

∴ Required Probability = 10/21

Step 1:

Let the envelope be denoted by E₁, E₂, E₃ and the corresponding letters by L₁, L₂, L₃

At least one letter should be in right envelope.

Let us consider all the favorable outcomes

Step 2:

(i) 1 letter in correct envelope and 2 in wrong envelope.

(ie) (E₁L₁,E₂L₃,E₃L₂),(E₁L₃,E₂L₂,E₃L₁),(E₁L₂,E₂L₁,E₃L₃)

(ii) Two letter in correct envelope.

(ie) (E₁L₁,E₂L₂,E₃L₃)

∴ No of favorable outcomes =4

Step 3:

Total no of outcomes =3!=3×2×1=6

∴ Required probability =n(E)/n(S)

⇒4/6

⇒2/3

Hence (A) is the correct answer.

Given:

Probability of Ramesh passing the government exam, P(A) = 8/10

Probability of Bala passing the government exam, P(B) = 6/8

Concept used :

Probability = number of favourable outcomes /total number of outcome

⇒ P(A) × P(B') + P(A') × P(B)

Calculation:

The probability that Ramesh and Bala fails is

⇒ P(A') fails = 2/10

⇒ P(B') fails = 2/8

Probability that any one fail = P(A) × P(B') + P(A') × P(B)

⇒ (8/10) × (2/8) + (2/10) × (6/8) = 1/5 + 3/20

⇒ 7/20

∴ The probability that Ramesh and Bala fails is 7/20

Calculation:

Number of possible outcomes when four dice are rolled = 6⁴

Total Possible outcomes where none of the dice shows 2 = 5⁴

Therefore, the number of outcomes in which at least one die shows 2 = 6⁴ - 5⁴ = 671

Given:

Black balls = 13

White balls = 12

Red balls = 15

Formula used:

ⁿC_r = n!/((n-r)! × r!)

Calculation:

Let S be the sample space then,

n(S) = Number of ways of selecting 4 balls out of 40 ball

⇒ n(S) = ⁴⁰C₄ = 40!/((40 – 4)! × 4!)

⇒ (40!/36! × 4!) = (40 × 39 × 38 × 37)/(4 × 3 × 2)

Let E be the event of getting 2 black and 2 red ball

n(E) = ¹³C₂ × ¹⁵C₂

⇒ (13 × 12)/2 × (15 × 14)/2

P(E) = n(E)/n(S)

⇒ P(E) = \(\frac{{\frac{{13{\rm{\;}} × {\rm{\;}}12}}{2} × \frac{{{\rm{\;}}\left( {15{\rm{\;}} × {\rm{\;}}14} \right)}}{2}}}{{\frac{{40{\rm{\;}} × {\rm{\;}}39{\rm{\;}} × {\rm{\;}}38{\rm{\;}} × {\rm{\;}}37}}{{4{\rm{\;}} × {\rm{\;}}3{\rm{\;}} × {\rm{\;}}2}}}}\)

⇒ (13 × 12 × 15 ×14 × 3 × 2)/(40 × 39 × 38 × 37)

⇒ 63/703

 ∴ The probability of getting 2 black ball and 2 red ball is 63/703

Concept: 

General Rule:

• The number of ways for selecting r from a group of n (n > r) = nCr 
• The probability of particular case = \(\rm \text{Number of ways for the case can be executed}\over{\text{Total number of ways for selection}}\)
• The sum of the possible case probabilities = 1

Calculation:

According to the question;

Probability of getting '1' = \(1\over2\)

Probability of getting '3' = \(1\over4\)

Probability of getting '5' = \(1\over6\)

Let the probability of getting an even number be x

As ∑ P = 1

⇒ x + \(1\over2\) + \(1\over4\) + \(1\over6\) = 1

⇒ x = \(\boldsymbol{1\over12}\)

Concept: 

General Rule:

• The number of ways for selecting r from a group of n (n > r) = nCr 
• The probability of particular case = \(\rm \text{Number of ways for the case can be executed}\over{\text{Total number of ways for selection}}\)

Bayes' theorem of probability: If E₁, E₂,…,E_n be a set of events in sample S. Let A be any event associated with S, then according to Bayes theorem,

 \(\rm P(E_i|A) = {P(E_i)P(A|E_i)\over\sum _{k=1}^nP(E_k)P(A|E_k)}\)

Where P(E_i) is the probability of Ei and P(E_i|A) is the probability that E_i assuming A has already occurred

Calculation:

In the given case 

Case A: The person is reported 2 

Case E₁: The number on dice is 2

∴ P(E₁) = \(1\over6\)

Case E2: The number on dice is not 2

∴ P(E2) = \(5\over6\)

Man reported 2 and 2 appears , i.e., man speaking true

⇒ P(A|E1) = \(3\over4\)

Man reported 2 and 2 do not appears, i.e., man speaking false

⇒ P(A|E2) = \(1\over4\)

For 2 actually appeared  

P(E₁|A) = \(\rm P(A|E_1)P(E_1)\over P(A|E_1)P(E_1) + P(A|E_2) P(E_2)\)

⇒ P(E1|A) = \(\rm {3\over4}\times{1\over6}\over{3\over4}\times{1\over6} + {1\over4}\times{5\over6}\)

⇒ P(E1|A) = \(\boldsymbol{3\over8}\)

Concept: 

General Rule:

• The number of ways for selecting r from a group of n (n > r) = nCr 
• The probability of particular case = \(\rm \text{Number of ways for the case can be executed}\over{\text{Total number of ways for selection}}\)

Calculation:

The total letters in box III = 3 + 1 + 2 + 5 + 3 + 4 = 18

Out of which:

White letters = 3 + 5 = 8

Blue letters = 1 + 3 = 4

Yellow letters = 2 + 4 = 6

Ways of selecting 2 letters out of 18 without replacement = ¹⁸C₂

Ways of selecting 2 letters out of 10 not white letters without replacement = 10C2  

Possibility of the letters picked are not white = \(\rm ^{10}C_2\over^{18}C_2\) = \(5\over 17\)

Concept: 

General Rule:

• The number of ways for selecting r from a group of n (n > r) = nCr 
• The probability of particular case = \(\rm \text{Number of ways for the case can be executed}\over{\text{Total number of ways for selection}}\)

For any two independent events A and B, if P(A) and P(B) is their probability of occurring then:

• P(A∩B) = P(A) × P(B)
• P(A∪B) = P(A) + P(B) - P(A∩B)
• P(A') = 1 - P(A)
• P(B') = 1 - P(B)

Conditional Probability:

If there are two cases A and B having a probability of P(A) and P(B), then,

Probability of A happens given that B definitely has happened P(A|B) = \(\rm P(A\;∩\; B)\over P( B)\)

Calculation:

In the given case 

Case A: Selecting the bag and there are 2 bags having an equal possibility of getting choose 

∴ P(A) = \(1\over2 \)

Case B: Selection of the green shirt from any bag

∴ P(B) = \({4\over 10} + {2\over 5}\) = \(4\over 5\)

Now P(A ∩ B) = P(A) × P(B)

⇒ P(A ∩ B) = \(1\over2 \) × \(4\over 5\) = \(2\over 5\)

Probability of case A happens given that case B definitely has happened

P(A|B) = \(\rm P(A\;∩ \;B)\over P( B)\)

⇒ P(A|B) = \({2\over5}\over{4\over5}\) = \(\boldsymbol{1\over2 }\)

Concept:

For any two independent events A and B, if P(A) and P(B) are their probability of occurring then:

• P(A ∩ B) = P(A) × P(B)
• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
• P(A') = 1 - P(A)
• P(B') = 1 - P(B)

Calculation:

Given P(A) = 0.6 and P(B) = 0.2

P(A ∩ B) = P(A) × P(B)

⇒ P(A ∩ B) = 0.6 × 0.2 = 0.12

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒ P(A ∪ B) = 0.6 + 0.2 - 0.12

⇒ P(A ∪ B) = 0.68

P((A ∪ B)') = 1 -  P(A ∪ B)

⇒ P((A ∪ B)') = 1 -  0.68

⇒ P((A ∪ B)') = 0.32

Concept:

P(A/B) = P(A∩B) / P(B)

\(\rm \overline{ P(A ∪B)}\)=\(\rm { P(\bar A ∩ \bar B)}\rm= 1-{ P(A ∪ B)}\)

P(A ∪ B) = P(A) +P(B) - P(A ∪ B)

Calculation: 

Here, P(G) = 2/5, P(Q) = 3/8 and P(G|Q) = 2/3

P(G|Q) = P(G ∩ Q) / P(Q) = 2/3

⇒ P(G ∩ Q) = 2/3 × 3/8 = 1/4

\(\rm P(\frac{\bar G}{\bar Q} )=\frac{P(\bar G∩ \bar Q)}{P(\bar Q)}=\frac{\overline {P( G∪ Q) }}{P(\bar Q)}\)

\(\rm =\frac{1-P(G∪ Q)}{P(\bar Q)}\)

P(Q̅) = 1 - P(Q) = 1-3/8 = 5/8

P(G ∪ Q) = P(G) + P(Q) - P(G∩Q)

= 2/5 + 3/8 - 1/4

= 21/40

\(\rm P(\frac{\bar G}{\bar Q} )=\frac{1-\frac {21}{40}}{\frac 58}\)

= 19/40 × 8/5

= 19/25

Hence, option (4) is correct.

Given:

Pink ball = 8

Green ball = 6 

Formula used:

ⁿC_r = n!/((n-r)! × r!)

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

where, n(E) → Required events, n(S) = Sample space

Calculation:

Let S be the sample space then,

n(S) be the number of ways to select 2 ball from 14 ball of the same colour 

n(S) = ¹⁴C₂ 

⇒ n(S) = (14 × 13)/2

⇒ 91

Let E be the event of getting 2 ball of same colour

n(E) = ⁸C₂ + ⁶C₂

⇒ (8 × 7)/2 + (6 × 5)/2

⇒ 28 + 15

⇒ 43

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

⇒ P(A) = 43/91

⇒ 43/91

∴ The probability of getting both balls of same colour is 43/91.

Given:

Total number of cards in two full deck cards = 2 × 52 = 104

Two cards are drawn at random.

Formula used:

Probability = Favorable outcome / Total outcomes

Calculation:

Total number of cards after the removal of all face cards = (104 – 2 × 4 × 3) = 80 [Each deck contains 4 × 3 = 12 face cards]

Total number of odd numbered cards = 40

Required probability = \({}_2^{40}C/{}_2^{80}C\) = (40 × 39)/(80 × 79) = 39/158

Given:

Number of tickets = 3500

Formula Used:

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

Calculation:

Total numbers of tickets sold = 3500 

Lucky draw prizes  = 100

Probability of getting a price  = 100/3500 = 1/35

∴ The probability of getting a prize if you buy 1 ticket is 1/35

Given:

A coin is tossed 3 times.

Formula used:

P (getting all heads) = (Number of favorable outcomes)/(Total number if outcomes)

Calculation:

Total possible outcomes

⇒ (TTT), (TTH), (THT), (HTT), (THH), (HTH), (HHT), (HHH)

The total number of outcomes = 8.

Number of favorable outcomes = (HHH)

P (getting all heads) = (Number of favorable outcomes)/(Total number if outcomes)

⇒ P (getting all heads) = 1/8

∴ The probability of getting all heads are 1/8.

Calculation:

While tossing a coin there are 2 possible outcomes (H or T), so for every toss possible cases are multiples by factor two.

Possible cases = [(H or T), (H or T), (H or T), (H or T), (H or T), (H or T), (H or T), (H or T) , i.e.,

Total cases = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256 or 2⁸

GIVEN :

A dice is rolled and at the same time 2 coins are tossed.

CONCEPT :

Probability.

FORMULA USED :

\(\rm P(H) = \dfrac{Number \ of \ Favourable \ Outcomes}{Total \ Number \ of \ Outcomes}\)

CALCULATION :

Probability of getting a number more than 3

⇒ Number of ways a number more than 3 can come = {4 , 5 , 6} = 3

Total number of possible outcome = {1,2,3,4,5,6} = 6

∴ P(A) = (Number of ways A can occur)/(Total number of possible outcomes)

⇒ Probability = 3/6 = 1/2

Probability of getting atleast one head

⇒ Number of ways atleast one head can occur = {(H,T) (T,H) (H,H)} = 3

Total number of possible outcomes = {(H,T) (T,H) (H,H) (T,T)} = 4

∴ Probability = 3/4

∴ The probability of getting a number more than 3 and atleast one head

⇒ (1/2) × (3/4) = 3/8

∴ The probability of getting a number more than 3 and atleast one head is 3/8

Given: 

Three dice are thrown simultaneously

Formula Used: 

Probability = (Possible outcomes/Total outcomes)

Calculation:

We know when 3 dices rolled together,

⇒ Total outcomes = 6³ = 216

Number of possible outcome when getting a sum of 3

⇒ (1, 1, 1) = 1 

Probability = 1/216 

∴  The probability of getting a sum of 3 is 1/216

Given:

Two dice are thrown simultaneously

Calculation:

When two dice are thrown simultaneously,

Number of possible outcomes are 36.

If getting the same number on both dice is taken as event,

Then for the 1^st event, 

⇒ Number of outcomes are 6.

For 2^nd event,

⇒ Number of outcomes/total number of possible outcomes

⇒ 6/(6 × 6)

⇒ 1/6

∴ The probability of getting the same number of both the dice is 1/6.

Formula used:

ⁿC_r = n!/{(n-r)! × r!}

Calculation:

Let S be the sample space then,

n(S) = Number of way of selecting 2 balls out of 31 ball

⇒ ³¹C₂ = 31!/[(31– 2)! × 2!]

⇒ (31 × 30)/2

⇒ 465

Let E be the event of getting both white balls

N(E) = ⁹C₂ = (9 × 8)/2

⇒ 36

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

⇒ 36/465

⇒ 12/155

∴ Probability of getting 2 white ball is 12/155

Formula used:

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

Where, n(S) → sample space, n(E) → favourable event

Calculation:

Let S be the sample space

S = 1, 2, 3, 4, 5, 6

N(S) = 6

Let E be the event of getting 5

⇒ n(E) = 1

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

⇒ 1/6

∴ Probability of a getting 5 up is 1/6

Formula used:

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

Calculation:   

Let S be the sample space                                                                     

Here S = HH, HT, TH, TT

⇒ n(S) = 4

Let E is event to getting bot are tails = TT

⇒ n(E) = 1

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

⇒ 1/4

∴ Probability of both are tails is 1/4

Formula used:

ⁿC_r = n!/((n-r)! × r!)

Calculation:

Let S be the sample space then,

n(S) = number of way of selecting 1 card out of 52 card

⇒ ⁵²C₁ = 52

Let E be the event of getting 9 of heart

Among 52 cards there is only one 9 of heart

n(E) = 1

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

⇒ 1/52

∴ Probability of getting 9 of heart is 1/52

Formula used:

P(E) = (Favorable events)/(Total events)

Calculations:

Total possible events = Total number of balls = 6 + 4 + 7 = 17

Favorable events = Total number of yellow balls = 4

P(E) = 4/17

∴ The probability of it to being a Yellow ball is 4/17

Formula used:

P(E) = (Favorable events)/(Total events)

Calculation

Total red card = 13 heart + 13 diamond = 26

King = 4

Total favorable events = The card is either red card or a king = 26 + 4 – 2 = 28

P(E) = 28/52 = 7/13

∴ The required probability is 7/13 

Formula used:

P(E) = (Favorable events)/(Total events)

Calculations:

In two throws of a dice = n(S) = 6 × 6 = 36

Events = {(6,2) (5,3) (4, 4) (2,6) (3,5)}

P(E) = 5/36

∴ The probability of getting a sum of 8 from two thrown throws of a dice is 5/36

Given:

18 numbers = 1 to 18

Calculation:

Number of odds in 1 to 18 = 9

⇒ Probability of picking out odd in first draw = 9/18

⇒ Probability of picking out odd in second draw = 8/17

⇒ Required probability = (1/2) × (8/17)

Given:

Cards = 52

Formula used:-

P(E) = Number of favorable outcomes/Total outcomes

Calculation:

Total no of favorable outcome is 4 out of 52 cards.

⇒ Total no of outcome is 52

⇒ Probability of getting “a jack” = 4/52

∴ Probability of getting a jack is 1/13

Concept:

P(E) = Probability of an event E = \(\rm \frac{\text{Favourable outcome}}{\text{Total outcome }} \)

Number of ways of choosing m objects by n people with no restriction = \(\rm m^{n}\)

Calculation:

Given

number of house (m) = 3

number of people applying for house(n) = 3

total number of way three people can apply for house = \(3^{3}\) = 27

Number of way in which all people apply for same house = 3

P(all three applying for same house) = \(\frac{3}{27}\) = \(\frac{1}{9}\) 

Calculation:

Let the number of green balls and blue balls present in the bag be 'g' and 'b' respectively

Also, number of red balls = 17

Total number of balls = (17 + g + b)

Probability of drawing a red ball = 17/(17 + g + b)

Probability of drawing a green ball = g/(17 + g + b)

Probability of drawing a blue ball = b/(17 + g + b)

Now a/q we have

⇒ g/(17 + g + b) = 1/7 + 17/(17 + g + b)

⇒ (g - 17)/(17 + g + b) = 1/7 

⇒ 6g - b = 136    .....(i)

also, 

⇒ b/(17 + g + b) = 1/21 + 17/(17 + g + b) 

⇒ (b - 17)/(17 + g + b) = 1/21

⇒ 20b - g = 374    .....(ii)    

using equation (i) and (ii), we get 

g = 26, b = 20

∴ Required total number of balls in the bag = (26 + 20 + 17) = 63

Concept :

Sure event: A sure event is an event, which always happens. The probability of a sure event is always 1.

Impossible event: An impossible event is an event that cannot happen. The probability of an Impossible event is always 0.

Calculation:

Using the above Concepts we can say,

Statements I is correct

Statements II is incorrect.

The correct option is 3 i.e. Statements I is correct but statement II is incorrect.

Formula used:

ⁿC_r = n!/((n-r)! × r!)

Calculation:

Let S be the sample space then,

n(S) = number of way of selecting 1 card out of 52 card

⇒ n(S) = ⁵²C₁ = 52

Let E be the event of getting a diamond card

Total number of diamond card = 13

n(E) = ¹³C1

⇒ 13

P(E) = n(E)/n(S) 

⇒ P(E) = 13/52

⇒ 1/4

∴ The probability of a diamond card is 1/4

Correct option b. (20/33)  

Explanation:

Let S be the sample space and A be the event of a van leaving first. n(S) = 100 , n(A) = 30 

Probability of a van leaving first: P(A) = 30/100 = 3/10 

Let B be the event of a lorry leaving first. n(B) = 100 – 60 – 30 = 10 

Probability of a lorry leaving first: P(B) = 10/100 = 1/10 

If either a lorry or van had left first, then there would be 99 vehicles remaining, 60 of which are cars. Let T be the sample space and C be the event of a car leaving second. 

n(T) = 99, n(C) = 60 

Probability of a car leaving after a lorry or van has left: 

P(B) = 60/99 = 20/33.

Correct option a. (14 and 13)

Explanation:

• The median is the middle value, so to rewrite the list in order: 13, 13, 13, 13, 14, 14, 16, 18, 21

• There are nine numbers in the list, so the middle one will be (9 + 1)/2 =5 = 5th number, So the median is 14.

• The mode is the number that is repeated more often than any other, so 13 is the mode.

Correct option (c) (3/13)

Explanation:

Total number of cards, n(S) = 52 

Total number of face cards, n(E) = 12 

P(E) = n(E)/ n(S) = 12/52 = 3/13

Given:

Total number of cards is 52

Formula used:

P (E) = Number of favorable outcomes/ total number of outcomes

Calculation:

Let E is a probability of an event.

Total number of cards = 52

There are 26 red cards in each suit.

 So, number of favorable outcomes = 26

⇒ P (E) = 26/52

⇒ 1/2

∴ The required probability is 1/2

Formula used:

Probability of a case to happen = Number of cases/Total number of cases

Calculation:

Total number of cards = 52

Total number of hearts in a deck = 13

Total number of diamonds in a deck = 13

∴ Probability of getting a heart or diamond = (13 + 13)/52 = 1/2

Given:

Green towel = 5

Blue towel = 7

Red towel = 8

Formula:

Probability (E) = Favourable outcomes/Total possible outcomes

Note: here, Probability denoted by ‘E’

The formula used for combination is –

nCr = n!/(n – r)! × r!

Here,

n = Sample of items.

r = Items are taken at a time.

Calculation:

Favourable outcomes –

2 (out of 5)

5C2 = 5!/(5 – 2)! × 2!

⇒ 5!/3! × 2!

⇒ 5 × 4 × (3!/3!) × 2!

⇒ 5 × (4/2) × 1

⇒ 20/2

⇒ 10

∴ Favourable outcomes will be 10

Total possible outcomes –

2 (out of 20)

20C2 = 20!/(20 – 2)! × 2!

⇒ (20!/18!) × 2!

⇒ 20 × 19 × (18!/18!) × 2!

⇒ 20 × (19/2) × 1

⇒ 10 × 19

⇒ 190

∴ Total possible outcomes will be 190

Now,

Probability (E) = Favourable outcomes/Total possible outcomes

Probability (E) = 10/190

⇒ 1/19

∴ The Probability that both are green will be 1/19

Concept:

A 3-digit number cannot start with digit 0 

Calculations:

Since a 3-digit number cannot start with digit 0,

The hundredth place can have any of the 4 digits.

Now, the tens and units place can have all the 5 digits. T

Therefore, the total possible 3-digit numbers are 4 × 5 × 5, i.e., 100.

The total possible 3 digit numbers having all digits same = 4

Hence, P (3-digit number with same digits) =4/100 = 1/25

Here n(S)=(6*6)=36 

let E=event of getting a total more than 7 

={(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5 ),(6,6)} 

P(E)=n(E)/n(S)=15/36=5/12.