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(i) 

(ii)

It is not P.d as P(X) < 0 when x = 3

(iii) 

It is not P.d as ΣP(X) = .9 ≠1

(iv)

Given: Two dice are thrown.

We know there are 6 faces on a dice. Which contains (1, 2, 3, 4, 5, 6).

Here two dice are thrown, and then we have two faces of dice (one of each).

So, the total sample space will be 6² = 36

∴ The sample space is:

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

As the random variable X takes 0, 1, 2, 3, …, n is said to be binomial distribution having parameters n and p, if the probability is given by

P(X = r) = ⁿC_r p^r q^n-r, where q = 1 – p and r = 0, 1, 2, 3, …

Similarly, in case of tossing a coin 3 times, we have

n = 3 and X has the values 0, 1, 2, 3 with p = ½, q = ½.

Therefore, it is said that the experiment of tossing a coin three times have binomial distribution.

Given: A coin is tossed four times.

We know that, the coins is tossed four time, then the no. of samples

2⁴ = 16

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

∴ The sample space is {HHHH, TTTT, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, HTHT, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH}

We know that, in a Bindmial distribution,

(i) There are 2 outcomes for each trial

(ii) There is a fixed number of trials

(iii) The probability of success must be the same for all the trials.

When coin is tossed, possible outcomes are Head and Tail.

Since coin is tossed three times, we have fixed number of trials.

Also probability of Head and Tail in each trial is 1/2.

Thus given experiment is said to have binomial distribution.

Correct option is (B) 4

If two coins are tossed simultaneously then possible outcomes are S = {HH, HT, TH, TT}.

\(\therefore\) Total number of outcomes is n(S) = 4.

Correct option is  B) 4

Correct option is (C) 1/2

R & S are consonant and 0 & E are vowels in word ROSE.

Total number of letters in ROSE is n(S) = 4

Total vowel letters in ROSE is n(V) = 2

\(\therefore\) P(V) \(=\frac{n(V)}{n(S)}=\frac24=\frac12\)

Hence, the probability that the letter chosen is a vowel is \(\frac{1}{2}.\)

Correct option is  C) 1/2 

Correct option is: A) 0

Correct option is: D) 2/5

Correct option is: D) 0 and 1

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We have, the sample space of an experiment of rolling a die is S = {1,2,3,4,5,6} 

Here, A = {2,3,5} and B = {1, 3, 5} 

(i) ’A or B’=A ∪ B = {1, 2, 3, 5} 

(ii) ’A and B’=A ∩ B = {3, 5} 

(iii) ’not A’= A’ = S – A = {1, 4, 6} 

(iv) ’A but not B ’= A – B = {2}

Probability is the chance of occurring of a certain event when measured quantitatively.

Area of bigger circle = 3.14×50×50

= 7850

Area of smaller circle = 3.14×20×20

=1256

Now if the player wants to score 100 points in a single shot then he will have to hit inside the smaller circle

So P(scoring 100) = 1256/7850

\(\frac 19 \times100 \% = 11\frac 19\)

X can take the values 0, 1, 2 It is a Random variable because the experiment is a random experiment.

Total balls = 5 + 8 + 7 = 20

(i) P(white ball) = 7/20

P(not white) = 1 - 7/20 = 13/20

(ii) P(green or red) = (8 + 5)/20 = 13/20

P(neither green nor red) = 1 - (13/20)

= 7/20

When a coin is tossed three times, the sample space is given by 

S = [HHH, HHT, HTH, THT, THH, HTT, TTH, TTT] 

E = {HHH, HTT, THT, TTH}, F = {TTT, HTH, THH, HHT}

E ∩ F = ϕ

P(E) = \(\frac{4}{8}\) = \(\frac{1}{2}\), P(F) = \(\frac{4}{8}\) = \(\frac{1}{2}\), P(E ∩ F) = ϕ

P(E) . P(F) = \(\frac{1}{2}\) x \(\frac{1}{2}\) x \(\frac{1}{4}\) ≠ P(E ∩ F)

∴ E and F are not independent events.

As, out of 3 events A,B and C only one can happen at a time which means no event have anything common. 

∴ We can say that A , B and C are mutually exclusive events. 

By definition of mutually exclusive events we know that:

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) 

According to question one event must happen. 

This implies A or B or C is a sure event. 

∴ P(A ∪ B ∪ C) = 1 …Equation 1 

We need to find odd against C 

Given, 

Odd against A = \(\frac{8}{3}\) 

⇒ \(\frac{P{(\bar A)}}{P(A)} = \frac{8}{3}\)

⇒ \(\frac{1-P( A)}{P(A)} = \frac{8}{3}\)

⇒ 8 P(A) = 3 – 3 P(A) 

⇒ 11 P(A) = 3

∴ P(A) = \(\frac{3}{11}\) …Equation 2 

Similarly, we are given with: Odd against B = \(\frac{5}{2}\) 

⇒ \(\frac{P{(\bar B)}}{P(B)} = \frac{5}{2}\)

⇒ \(\frac{1-P( B)}{P(B)} = \frac{5}{2}\) 

⇒ 5 P(B) = 2 – 2 P(B) 

⇒ 7 P(B) = 2 

∴ P(B) = \(\frac{2}{7}\)…Equation 3

From equation 1,2 and 3 we get:

P(C) = 1- \(\frac{3}{11}-\frac{2}{7}=\frac{77-21-22}{77} = \frac{34}{77}\)  

∴ P(C’) = 1 – \(\Big(\frac{34}{77}\Big)=\frac{43}{77}\)  

∴ Odd against C = \(\frac{p{(\bar c)}}{p(c)}\) = \(\frac{43}{\frac{77}{\frac{34}{77}}}\) = \(\frac{43}{34}\)

Let A and B are two events. 

As, out of 2 events A and B only one can happen at a time which means no event have anything common. 

∴ We can say that A and B are mutually exclusive events. 

By definition of mutually exclusive events we know that: 

P(A ∪ B) = P(A) + P(B) 

According to question one event must happen. 

This implies A or B is a sure event. 

∴ P(A ∪ B) = P(A) + P(B) = 1 …Equation 1 

Given, P(A) = (2/3)P(B) 

To find: odds in favour of B

∴ P(B’) + \(\frac{2}{3}\) P(B) = 1 

⇒ \(\frac{5}{3}\) P(B) = 1 ⇒ P(B) = \(\frac{3}{5}\)

∴ P(B’) = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\)

∴ Odd in favour of B = \(\frac{P(B)}{P(\bar B)}\) = \(\frac{3}{\frac{5}{\frac{2}{5}}}\) = \(\frac{3}{2}\)

Total no. of outcomes = 52 {52 cards}

(i) E⟶ event of getting a black king

No of favourable outcomes = 2{king of spades & king of clubs}

We know that, P(E) = (No. of favorable outcomes)/(Total no.of possible outcomes) = 2/52 = 1/26

(ii) E⟶ event of getting either a black card or a king.

No. of favourable outcomes = 26 + 2 {13 spades, 13 clubs, king of hearts & diamonds}

P(E) = (26+2)/52 = 28/52 = 7/13

(iii) E⟶ event of getting black & a king.

No. of favourable outcomes = 2 {king of spades & clubs}

P(E) = 2/52 = 1/26

(iv) E⟶ event of getting a jack, queen or a king

No. of favourable outcomes = 4 + 4 + 4 = 12 {4 jacks, 4 queens & 4 kings}

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

(v) E⟶ event of getting neither a heart nor a king.

No. of favourable outcomes = 52 – 13 – 3 = 36 {since we have 13 hearts, 3 kings each of spades, clubs & diamonds}

P(E) = 36/52 = 9/13

(vi) E⟶ event of getting spade or an all.

No. of favourable outcomes = 13 + 3 = 16 {13 spades & 3 aces each of hearts, diamonds & clubs}

P(E) = 16/52 = 4/13

(vii) E⟶ event of getting neither an ace nor a king.

No. of favourable outcomes = 52 – 4 – 4 = 44 {Since we have 4 aces & 4 kings}

P(E) = 44/52 = 11/13

(viii) E⟶ event of getting neither a red card nor a queen.

No. of favourable outcomes = 52 – 26 – 2 = 24 {Since we have 26 red cards of hearts & diamonds & 2 queens each of heart & diamond}

P(E) = 24/52 = 6/13

(ix) E⟶ event of getting card other than an ace.

No. of favourable outcomes = 52 – 4 = 48 {Since we have 4 ace cards}

P(E) = 48/52 = 12/13

(x) E⟶ event of getting a ten.

No. of favourable outcomes = 4 {10 of spades, clubs, diamonds & hearts}

P(E) = 4/52=1/13

(xi) E⟶ event of getting a spade.

No. of favourable outcomes = 13 {13 spades}

P(E) = 13/52 = 1/24

(xii) E⟶ event of getting a black card.

No. of favourable outcomes = 26 {13 cards of spades & 13 cards of clubs}

P(E) = 26/52=1/2

(xiii) E⟶ event of getting 7 of clubs.

No. of favourable outcomes = 1 {7 of clubs}

P(E) = 1/52

(xiv) E⟶ event of getting a jack.

No. of favourable outcomes = 4 {4 jack cards}

P(E) = 4/52=1/13

(xv) E⟶ event of getting the ace of spades.

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

P(E) = 1/52

(xvi) E⟶ event of getting a queen.

No. of favourable outcomes = 4 {4 queens}

P(E) = 4/52 = 1/13

(xvii) E⟶ event of getting a heart.

No. of favourable outcomes = 13 {13 hearts}

P(E) = 13/52 = 1/4

(xviii) E⟶ event of getting a red card.

No. of favourable outcomes = 26 {13 hearts, 13 diamonds}

P(E) = 26/52 = 1/2

Total no. of possible outcomes = 8 {3 red, 5 black}

(i) E ⟶ event of getting red ball.

No. of favourable outcomes = 3 {3 red}

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

P(E) = 3/8

(ii) Bar E ⟶ event of getting no red ball.

()+P(Bar E)=1

P(̅)=1−P(Bar E)

= 1 − 3/8 = 5/8

Total no. of possible outcomes = 90 {1, 2, 3, … 90}

(i) E ⟶ event of getting 2 digit no.

No. of favourable outcomes = 81 {10, 11, 12, …. 90}

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

P(E) = 81/90 = 9/10

(ii) E ⟶ event of getting a perfect square.

No. of favourable outcomes = 9 {1, 4, 9, 16, 25, 26, 49, 64, 81}

P(E) = 9/90=1/10

(iii) E ⟶ event of getting a no. divisible by 5.

No. of favourable outcomes = 18 {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90}

P(E) = 18/90 = 1/5

(c) P(X) = P(Y) > P(Z) 

P(X) = \(\frac{26}{52}\) + \(\frac{4}{52}\) - \(\frac{2}{52}\) = \(\frac{28}{52}\) 

(∵ There are 26 black cards, 4 kings and 2 black kings)

P(Y) = \(\frac{13}{52}\) + \(\frac{13}{52}\) + \(\frac{4}{52}\) - \(\frac{1}{52}\) - \(\frac{1}{52}\) = \(\frac{28}{52}\)

(∵ There are 13 clubs, 13 hearts and 4 jacks, 1 jack of clubs and 1 jack of hearts)

P(Z) = \(\frac{4}{52}\) + \(\frac{13}{52}\) + \(\frac{4}{52}\) - \(\frac{1}{52}\) - \(\frac{1}{52}\) = \(\frac{19}{52}\)

(∵ There are 4 aces, 13 diamonds, 4 queens, 1 ace of diamond, 1 queen of diamond) 

∴ P(X) = P(Y) > P(Z).

Here, p = 1/6 + 1/6 + 1/6 = ½ ⇒ q = 1 – ½ = ½ and n = 5

Now,

P(x = r) = ⁿC_r p^r q^{n – r} = ⁵C₃ (1/2)³(1/2)^5-3

= [5!/(3!2!)].(1/2)³(1/2)² = 10.1/8.1/4 = 5/16

Therefore, the required probability is 5/16.