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Total no. of possible outcomes = 52 (52 cards)

(i) E ⟶ event of getting a king of red suit

No. of favourable outcomes = 2 {king heart & king of diamond}

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

(ii) E ⟶ event of getting face card

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

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

(iii) E ⟶ event of getting red face card

No. favourable outcomes = 6 { kings, queens, jacks of hearts & diamonds}

P(E) = 6/26 = 3/26

(iv) E ⟶ event of getting a queen of black suit

No. favourable outcomes = 6 { kings, queens, jacks of hearts & diamonds}

P(E) = 6/26 = 3/26

(v) E ⟶ event of getting red face card

No. favourable outcomes = 6 { queen of spades & clubs}

P(E) = 1/52

(vi) E ⟶ event of getting a spade

No. favourable outcomes = 13 {13 spades}

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

The odds against her husband living till he is 85 are 7 : 5. 

Let P(H’) = P(husband dies before he is 85) = 7/7+5 = 7/12

So, the probability that the husband would be alive till age 85 

P(H) = 1 – P(H’) = 1 – 7/12 = 5/12

Similarly, P(W’) = P(Wife dies before she is 81) 

Since the odds against wife will be alive till she is 81 are 5 : 3.

∴ P(W’) = 5/5+3 = 5/8

So, the probability that the wife would be alive till age 81 

P(W) = 1 – P(W’) = 1 – 5/8 = 3/8

(i) Required probability 

P(H ∪ W) = P(H) + P(W) – P(H ∩ W) 

Since H and W are independent events, P(H ∩ W) = P(H) . P(W) 

∴ Required probability = P(H) + P(W) – P(H) . P(W)

= \(\frac {5}{12} + \frac {3}{8} -\frac{5}{12}\times\frac{3}{8}\)

= 40+36-15/96

= 61/96

(ii) Required probability = P(H ∩ W’) + P(H’ ∩ W) 

Since H and W are independent events, H’ and W’ are also independent events. 

∴ Required probability = P(H) . P(W’) + P(H’) . P(W)

= \(\frac {5}{12} + \frac {5}{8}+\frac{7}{12}\times\frac{3}{8}\)

= 25+21/96

46/96

= 23/48

Correct option is: A) \(\frac 1{80}\)

A sample space consists of 80 elementary events that are equally likely.

\(\therefore\) Probability of each of them = \(\frac 1{80}\)

Correct option is: A) \(\frac{1}{80}\)

Correct option is: C) \(\frac{1}{3}\)

There are 3 sides in a right angled triangle in which only one side (longest side) is hypotenuse.

\(\therefore\) Probability that selected side is hypotenuse = \(\frac{1}{3}\)

Correct option is: C) \(\frac{1}{3}\)

Correct option is: A) 0.18

\(\because\) P (E) + P(\(\overline E\)) =1

\(\Rightarrow\) P (\(\overline E\)) = 1- P(E)

= 1- 0.82 (\(\because\) P(E) = 0.82 (Given))

= 0.18 

Correct option is: A) 0.18

Correct option is: D)\(\frac{1}{2^3}\)

When 3 coins are tossed, total possible outcomes are

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

\(\therefore\) Total No of possible outcomes when 3 coins are tossed 

= 8 = \(2^3\)

Fabourable outcome that favours to event of getting no head is {TTT}

\(\therefore\) Total fabourable outcomes = 1

\(\therefore\) Probability of getting no head = \(\frac {Total \,favourable\, outcomes}{Total \,No \,of \,possible \,outcomes}\) = \(\frac 18 = 2^{\frac 13}\)

Correct option is: D) \(\frac{1}{2^3}\)

Correct option is: B) \(\frac{1}{2}\)

The probability of an event cannot be

(ii) 3.8 i.e. the probability of an even cannot exceed 1.

(iv) i.e. -0.8 and

(vi) -2/5, This is because probability of an even can never be less than 1.

Total letter on the dice = 6

(i) Number of times A appears = 3

Therefore, numbers of favorable outcomes = 3

P (getting letter A) = 3/6 = 1/2

(ii) Number of times D appears = 1

Therefore, numbers of favorable outcomes = 1

P (getting letter D) = 1/6

There are 4 suits in a pack of 52 cards

13- Heart

13- Diamond

13- Spades

13- Clubs

Each suit has one ace, 1jack, 1king, 1queen

So total ace cards = 4

(i) The probability of getting an Ace = 4/52

(ii) There are 48 cards which are not ace

The probability of not getting an Ace = 48/52

Outcomes= (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) (6,3),(6,4),(6,5),(6,6),(3,6),(4,5),(4,6),(5,4),(5,5),(5,6)

P (sum greater than 8) = 10/36

(ii) Less than or equal to 12 = ?

Outcomes = All Outcomes have sum less than or equal to 12.

P (sum Less than or equal to 12) = 36/36 = 1

(iii) Sum equal to 7

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

P (sum equal to 7) = 6/36

(iv) Divisible by 3 or 4

Outcomes =

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

P(divisible by 3 or 4) = 20/36

P (E) = 0.05 given 

P (E) + P (not E) = 1 

P (not E) = 1 – 0.05 = 0.95

Outcomes (X) = 1,2,3,4,5,6

(i)The number 5 appers only once in a throw of dice So P(X)= 1/6

(ii) There are either chances of getting 3 or 4 P(X) = 2/6

(iii) Prime Number:- a number which is divisible by 1 or number itself.

So favourable cases are 2,3,5 P(X) = 3/6 

(iv) Number greater than 4 are 5,6 so P(X) = 2/5

(v) A Dice contains only n1,2,3,4,5,6 as the number, so there is no chance of getting a number than 6 So P(X)=0

(vi) Since A Dice contains only n1,2,3,4,5,6 as the number So Number, less than 6 is P(X) = 5/6

The correct option is: (B) 1/3

Explanation:

Total number of outcomes when two dice are thrown = 36

Let A be the event of getting a number always greater than 4 on second dice.

. ^.. A = {(1, 5), (1, 6), (2,5), (2,6), (3, 5), (3, 6), (4, 5), (4, 6), (5, s), (5, 6), (6, s), (6, 6)) 

.'. Number of possible outcomes = 12

. ^.. P(A) = 12/36 = 1/3

Probability of an event cannot be negative or more than one so -1.5 can never be probability of any event

Total number of possible outcomes, n(S) = 18 

(i) Number of favorable outcomes, 

n(E) = 14

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{14}{18}\) = \(\frac{7}{9}\)

(ii) Number of events of getting a black ball, 

n(E) = 4

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{18}\) = \(\frac{2}{9}\)

Probability of not getting a black ball = 1 – P(E)

= 1 - \(\frac{2}{9}\) = \(\frac{7}{9}\)

(iii) Number of favorable outcomes, 

n(E) = 8

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{8}{18}\) = \(\frac{4}{9}\)

(i) 0 Since there are only blue balls so taking, green balls will be zero.

(ii)1 Since there are only blue balls.

Total numbers of elementary events are: 7 

Let E be the event of having 53 Sundays 

Note- a non-leap year has 365 days or 52 weeks and 1 day. This one day could be any day amongst Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday. Out of these favorable event is one; Sunday. 

Number of favorable outcome is = 1 

P (Sunday) = P (E) = \(\frac{1}7\)

Total numbers of elementary events are: 6 x 6 = 36 

Let E be the event of getting a perfect square 

Favorable outcomes are: 4, 9 4 

and 9 can be rolled in combinations of: (2,2), (1,3), (3,1) , (3,6), (4,5), (5,4), (6,3) 

Numbers of favorable outcomes are = 7 

P (perfect square) = P (E) = \(\frac{7}{36}\)

Total numbers of elementary events are: 6 x 6 = 36 

Let E be the event of getting an even number on both the dice

Favorable outcomes are: (2,2) , (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6) 

Numbers of favorable outcomes are = 9 

P (getting even number on both dice) = P (E) = \(\frac{9}{36}\) = \(\frac{1}{4}\)

Total numbers of elementary events are: 52 

Let E be the event of drawing card other than ace 

Favorable numbers of events are: 52 – 4 = 48 

P (not drawing ace) = P (E) = \(\frac{48}{52}\) = \(\frac{24}{26}\) = \(\frac{12}{13}\)

Total numbers of elementary events are: 30 

Let E be the event of selecting a prime number 

Favorable outcomes are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 

Numbers of favorable outcomes are: 10 

P (prime number) = P (E) = \(\frac{10}{30}\) = \(\frac{1}{3}\)

Total numbers of elementary events are: 25 

Let E be the event of getting a prime number 

Favorable outcomes are: 2, 3, 5, 7, 11, 13, 17, 19, 23 

Numbers of favorable outcomes are = 9 

P (prime number) = P (E) = \(\frac{9}{25}\)

Outcomes = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25

Prime Numbers = 2,3,5,7,11,13,17,19,23

Let x be the event of getting prime numbers

P(x) = 9/25

Therefore the probability of getting non-prime number = 16/25

Total number of possible outcomes, n(S) = 25 

Number of events of getting a prime number, 

n(E) = 9

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{9}{25}\)

Number of events of not getting a prime number = 1 – P(E)

= 1 - \(\frac{9}{25}\) = \(\frac{16}{25}\)

Total number of outcomes = (HHH),(HHT),(HTH),(THH),(TTH),(TTT),(THT),(HTT)

(i) P(exactly two tails) = 3/8

(ii) P(at least one head) = 7/8

(iii) P(at least one head and one tail) = 1

Total number of possible outcomes, n(S) = 20 

(i) Number of favorable outcomes, 

n(E) = 13

∴P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{13}{20}\)

(ii) Number of events of getting a black ball, 

n(E) = 7

∴P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{7}{20}\)

Probability of not getting a black ball = 1 – P(E)

= 1 - \(\frac{7}{20}\) = \(\frac{13}{20}\)

(iii) Number of favorable outcomes, 

n(E) = 5

∴P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{5}{20}\) = \(\frac{1}{4}\)

Total number of possible outcomes, n(S) = 30 

Number of favorable outcomes of selecting a card divisible by 3, 

n(E) = 10

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{10}{30}\) = \(\frac{1}{3}\)

Number of favorable outcomes of not selecting a card divisible by 3 = 1 – P(E)

= 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)

(i) Total number of cards = 30

Number divisible by 3 = 3,6,9,12,15,18,21,24,27,30

P(probability that the number on the selected cards is not divisible by 3) = 20/30

(ii) P(E) = 0.8

P(not E) = 1-0.08 =0.92

(iii) P (the probability of the 2 students passing the examination) = 1 - 0.895 = 0.105

Total cards = 30

Number divisible by 3 = 3,6,9,12,15,18,21,24,27,30

= Total number of favourable outcomes 30 - 10 = 20

.^. . Required probability = 20/30 = 2/3

The correct option is: (D) 3/100

Explanation:

Total number of possible outcomes = 100 

Numbers from 1 to 100 which are divisible by 9 and perfect square are 9, 36 and 81. 

Number of favourable outcomes = 3

. ^..  Required probability = 3/100

Total numbers of elementary events are: 3+ 4 + 2= 9 

Let E be the event of getting not an orange marble 

Numbers of Favorable outcomes are: 3 + 4 = 7 

P (not an orange marble) = P (E) = \(\frac{7}9\)

Total numbers of elementary events are: 10 

Let E be the event of getting selected number being the average

Favorable outcome is = \(\frac{70}{10}\) = 7 

Average = total of observations/ numbers of observation 

Total of observations = 3+ 5+ 5+ 7+ 7+ 7+ 9+ 9 + 9+ 9 = 70 

Numbers of observation = 7 

Numbers of favorable outcomes are: 3 

P ( selecting average number) = P (E) = \(\frac{3}{10}\)

(i) The probability of an impossible event is zero.

(ii) The probability of a sure event is one.

(iii) For any event E, P(E) + P(not E) = 1.

(iv) The probability of a possible but not a sure event lies between zero and one.

(v) The sum of probabilities of all the outcomes of an experiment is one.

When two coins are tossed simultaneously then the possible outcomes obtained are {HH, HT, TH, and TT}.

Here H denotes head and T denotes tail.

Therefore, a total of 4 outcomes obtained on tossing two coins simultaneously.

Number of favourable outcome(s) for no head is {TT}

Number of favourable outcome(s) for one head is {HT, TH}

Therefore, the event of getting at most one head has 3 favourable outcomes. These are TT, HT and TH.

Therefore, the probability of getting at most one head = 3/4.