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We know that,

When a coin is tossed, there are only two possible OUTCOMES, either heads or tails

We know that the sample SPACE S = {HH, HT, TH, TT}

The EVENT E that at least one of them is head = {HH, HT, TH}

Probability P(E) = n(e)/n(s) = 3/4

<P>We know that,

The vowels in ENGLISH Alphabets are: a, e, i, o, u

The sample Space S = {1M, 4I, 4S, 2P}

The event E = {4I}

Probability P(E) = n(e)/n(s) = 4/11

On THROWING TWO DICE, n(S) = 6 × 6 = 36

Number of events of getting a total of 7, n(E) = 6, i.e. (1,6), (2,5) , (3,4), (4,3) , (5,2), (6,1)

⇒ Here, S = {(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)}

⇒ n(S) = 2³ = 8

⇒ Let event E = getting at least two tails

⇒ {(HTT), (THT), (TTH), (TTT)}}

∴ n(E) = 4

Now, probability of getting at least two tails

Let S be the sample space.

Then, total number of BALLS = (4 + 6 + 8 + 10) = 28 balls

Thus, N(S) = 28

Let E = event that the ball drawn is black

? There are 10 black balls in the bag

Thus, n(E) = 10

∴ PROBABILITY of occurrence of event;

PROBABILITY of choosing 1ST basket = ½

∴ Probability of choosing BLUE balls from 1^st basket = ½ × 11/14 = 11/28

Probability of choosing 2^nd basket = ½

∴ probability of choosing blue balls from 2nd basket = ½ × 5/14 = 5/28

∴ Probability of drawing a blue BALL,

⇒ 11/28 + 5/28

⇒ 16/28

Let the number of BOYS be x

Total number of people in group = 25 + x

Probability of choosing a boy from a group = x/(25 + x)

x/(25 + x) = 3/8 ⇒ x = 15

Here, at least one of the two cards needs to be red.

This can be interpreted as: the required cases contain all the cases EXCEPT when NONE of the cards drawn is red.

We know that, there are 26 red and 26 black cards in a regular deck.

Let E₃ be the event that none of the cards drawn is red, i.e. both the cards are black.

∴ NUMBER of ways to draw 2 black cards out of 26 black cards = ²⁶C₂

⇒ PROBABILITY that none of the cards is red, P(E₃) = ²⁶C₂/⁵²C₂ = 25/102

Numbers from 1 to 33 divisible by 5 = {5, 10, 15, 20, 25, 30}

Numbers from 1 to 33, not divisible by 3 but divisible by 5 = {5, 10, 20, 25}

Number of possible outcomes = 33

∴ Probability = $(\FRAC{{n\left( {Favourable\;Events} \right)}}{{n\left( {Possible\;outcomes} \right)}}\; = \;\frac{4}{{33}})$

Total number of balls in the box = 3 + 5 = 8

? A ball is drawn, Sample SPACE = N(S) = ⁸C₁ = 8

First we will FIND QUANTITY A,

Quantity A: Probability that the ball is black.

Quantity A = n(A)/n(S) = ³C₁/8 = 3/8

Now,

Quantity B: Probability that the ball is white.

Quantity B = n(B)/n(S) = ⁵C₁/8 = 5/8

∴ Quantity B > Quantity A

Number of not DEFECTIVE protractors = 164 - 21 = 143

Total number of protractors = 164

∴ PROBABILITY that RUHI BUYS the protractor = $(\frac{{143}}{{164}} = \frac{{286}}{{328}})$

<P>Total number of caps in box = (2 blue + 4 red + 5 green + 1 yellow) caps = 12

Let,

E1 = event of picking a blue cap

E2 = event of picking a yellow cap

Then,

P(E1 ) = 2/12

P(E2) = 1/12

? Probability = [No of Favorable Outcomes /No. of Total Outcomes]

⇒ No. of Total Outcomes = Total Ways of Arranging word "PROBABILITY"

NUMBER of ways of arranging ‘n’ things in which ‘p’ are of one type ‘q’ are of one type and rest are different = $(\frac{{n!}}{{p! \times q!}})$

= Arranging a 11 letter word where TWO letters occur TWICE

$(= \frac{{11!}}{{2! \times 2!}} = \frac{{11!}}{4})$

Assume two B as a one unit and two ‘I’ as another unit, So Number of favourable cases in which two ‘B’ together and 2 ‘I’ are together = 9!

⇒ Probability $(= \frac{{9!}}{{\frac{{11!}}{4}}})$

$(= \frac{4}{{10 \times 11}})$

Numbers from 5 to 55 divisible by 7 = {7, 14, 21, 28, 35, 42, 49}

Numbers from 5 to 55, not divisible by 3 but divisible by 7 = {7, 14, 28, 35, 49}

Number of total possible OUTCOMES = 51

Statement I:

Total no. of balls = 10 + 12 + 20 = 42

Number of red and BLUE balls = 10 + 12 = 22

∴ The SUM of probabilities of finding a red and blue balls = 22/42 = 11/21

∴ Quantity A = 11/21 ≈ 0.52

Statement II:

Probability of getting a head = 1/2

∴ Probability of getting two heads = 1/2 × 1/2 = 1/4

∴ Quantity B = 1/4 = 0.25

? 0.52 > 0.25

P(A) = $(\FRAC{1}{4})$

P(B) = $(\frac{1}{2})$

$(P\left( {A \cap B} \right) = \frac{1}{8})$

P(A or B) = P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒ Probability of selection of a ball from bag 1 = P(B₁) = 1/2

⇒ Probability of selection of a ball from bag 2 = P(B₂) = 1/2

Now, there are total 11 balls in the bag 1 (4 red balls + 7 blue balls),

⇒ Probability of getting red ball from bag 1 = P(R/B₁) = number of red balls/total balls = 4/11

⇒ Probability of getting blue ball from bag 1 = P(B/B₁) = number of blue balls/total balls = 7/11

Also, there are total 8 balls in the bag 2 (5 red balls + 3 blue balls),

⇒ Probability of getting red ball from bag 2 = P(R/B₂) = number of red balls/total balls = 5/8

⇒ Probability of getting blue ball from bag 2 = P(B/B₂) = number of blue balls/total balls = 3/8

Quantity A:

It is given that ball DRAWN is red, we have to find the probability of drawn ball being from bag 2,

So, using Baye’s theorem,

⇒ P(B₂/R) = {P(B₂) × P(R/B₂)}/[{P(B₂) × P(R/B₂)} + {P(B₁) × P(R/B₁)}]

⇒ P(B₂/R) = {(1/2) × (5/8)} / [{(1/2) × (5/8)} + {(1/2) × (4/11)

$(\Rightarrow {\rm{\;P}}\left( {{{\rm{B}}_2}/{\rm{R}}} \right) = \FRAC{{{\rm{P}}\left( {{{\rm{B}}_2}} \right) \times {\rm{P}}\left( {\frac{{\rm{R}}}{{{{\rm{B}}_2}}}} \right)}}{{{\rm{P}}\left( {{{\rm{B}}_2}} \right) \times {\rm{P}}\left( {\frac{{\rm{R}}}{{{{\rm{B}}_2}}}} \right) + \;{\rm{P}}\left( {{{\rm{B}}_1}} \right) \times {\rm{P}}\left( {\frac{{\rm{R}}}{{{{\rm{B}}_1}}}} \right)}})$

Putting all the values in the formula, we get

$(\Rightarrow {\rm{P}}\left( {{{\rm{B}}_2}/{\rm{R}}} \right) = {\rm{\;}}\frac{{\frac{1}{{2{\rm{\;}}}} \times \frac{5}{8}}}{{\frac{1}{2} \times \frac{5}{8}\; + \;\frac{1}{2} \times \frac{4}{{11}}}} = \frac{{\frac{5}{{16}}}}{{\frac{5}{{16}}\; + \;\frac{4}{{22}}}} = 110/174{\rm{\;}} = {\rm{\;}}55/87)$

⇒ Quantity A = 55/87

Quantity B:

It is given that ball drawn is blue, we have to find the probability of drawn ball being from bag 1,

So, using Baye’s theorem,

$(\Rightarrow {\rm{\;P}}\left( {\frac{{{{\rm{B}}_1}}}{{\rm{B}}}} \right) = \frac{{{\rm{P}}\left( {{{\rm{B}}_1}} \right) \times {\rm{\;P}}\left( {\frac{{\rm{B}}}{{{{\rm{B}}_1}}}} \right)}}{{{\rm{P}}\left( {{{\rm{B}}_1}} \right) \times {\rm{\;P}}\left( {\frac{{\rm{B}}}{{{{\rm{B}}_1}}}} \right) + \;{\rm{P}}\left( {{{\rm{B}}_2}} \right) \times {\rm{\;P}}\left( {\frac{{\rm{B}}}{{{{\rm{B}}_2}}}} \right)}})$

Putting all the values in the formula,

$(\Rightarrow {\rm{\;P}}\left( {\frac{{{{\rm{B}}_1}}}{{\rm{B}}}} \right) = {\rm{\;}}\frac{{\frac{1}{2} \times \frac{7}{{11}}}}{{\frac{1}{2} \times \frac{7}{{11}}\; + \;\frac{1}{2} \times \frac{3}{8}}} = \frac{{\frac{7}{{22}}}}{{\frac{7}{{22}}\; + \;\frac{3}{{16}}}}{\rm{\;}} = 112/178{\rm{\;}} = {\rm{\;}}56/89)$

⇒ Quantity B = 56/89

PROBABILITY of passing 1^st paper = 3/5 

Probability of passing 2^nd paper = 7/10

Probability of passing 3^rd paper = 7/10 

∴ Probability that Prakash PASSES in all the three papers, 

= 3/5 × 7/10 × 7/10

= 147/500 

= 0.294

Number of possible OUTCOMES = n(S) = 47

Number of favourable outcomes = 47[No number is divisible BYBOTH 11 and 13 in between 153 and 199]

There are total 20 people in the party.

∴ Total number of possible shake HANDS = 20c₂ = 190

Number of ways where an woman shakes hand with her HUSBAND = 10 [There are 10 women]

Number of ways where Indian women shakes hand with AMERICAN MAN = 5 × 5 = 25

∴ Total number of ways where no wife shakes hand with her own husband and no Indian woman shakes hand with an American man = (190 – 25 – 10) = 155

⇒ Required probability = (155/190) = 31/38