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The correct option is: (A) 12 

Explanation:

Let there be b blue, g green and w white marbles in the jar. Then, b + g + w = 54                ....(i)

. ^.. P (Selecting a blue marble) = b/54

It is given that the probability of selecting a blue marble is 1/3.

. ^.. 1/3 = b/54 => b = 18

We have, P (Selecting a green marble) = 4/9 (given)

=> g/54 = 4/9 => g = 24

Substituting the values of b and g in (i), we get 

18 + 24 + w = 54 + w = 12 

Hence, the jar contains 12 white marbles. 

1. (d) 3/5

2. (b) 1

3. (a) 11/15

4. (c) 1

5. (c) 9/11

Give that bag I contains 8 white and 7 black balls and Bag II contain 5 white and 4 black balls. 

One ball from I bag randomly put in II bag. 

So the possibility is that ball taken out from bag I is white. 

Then probability that white ball is choosen from bag I = 8/15

Now total number of balls in bag II = 5 + 1 = 6

Probability that white ball is choosen from bag II = 6/10

When both events occurs together 

∴ Probability = 8/15 × 6/10 = 48/150

Another possibility is that ball is choosen from bag I is black. 

Then probability that black ball is choosen from bag = 7/15

Now number of black ball in bag II = 4 + 1 = 5 

∴ Probability that black ball is choosen = 5/10

Probability that both events happen together 

= 7/15 × 5/10 = 35/150

∵ Both events are mutually exclusive so only one event can happen. 

∴ Required probability = 48/150 + 35/120

= 83/150

(i) Sum of probabilities = 0.4 + 0.4 + 0.2 = 1 

Hence, given distribution is a probability distribution. 

(ii) Sum of probabilities = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1 

Hence, the given distribution is not a probability distribution. 

(iii) Here, one of the probability is – 0.1. which is negative. 

Hence, this distribution is not probability distribution.

Total number of ways of putting three letters into three envelopes is 

3P₃ = 3! (ways) = 6. 

The number of ways in which none of the letter is put into proper envelope is 2.

∴ P(none) = 2/6 = 1/3

∴ p(atleast one letter is in its proper envelop) = 1 - p (none)

= 1 - 1/3 = 2/3

Total number of tickets sold = 10,000

Number prizes awarded = 10
(i) If we buy one ticket, then

P (getting a prize) = 10/10000=1/1000

∴ P (not getting a prize) =  1-1/1000 = 999/1000

(ii) If we buy two tickets, then
Number of tickets not awarded = 10,000 − 10 = 9990

P (not getting a prize) = ⁹⁹⁹⁰C₂ / ¹⁰⁰⁰⁰C₂

(iii) If we buy 10 tickets, then
P (not getting a prize) = ⁹⁹⁹⁰C₁₀ / ¹⁰⁰⁰⁰C₁₀

Correct option is: C) unlikely

Buying a lottery ticket and win the Jackpot is unlikely event.

Correct option is: C) unlikely

Total numbers of elementary events are = 9 

Let E be the event of getting an odd number 

The favorable outcomes are: 1, 3, 5, 7, 9 

Numbers of favorable events= 5 

P (odd number) = P (E) = \(\frac{5}9\)

For each event to be a valid assignment of probability, the probability of each event in sample space should be less than 1 and the sum of probability of all the events should be exactly equal to 1 

(i) it is valid as each P(w_i) (for i=1 to 7) lies between 0 to 1 and sum of P(w₁) = 1 

(ii) it is valid as each P(w_i) (for i = 1 to 7) lies between 0 to 1 and sum of P(w₁) =1 

(iii) it is not valid as sum of P(w_i)=2.8 which is greater than 1 

(iv) it is not valid as P(w₇) = \(\frac{15}{14}\)  which is greater than 1.

We know that, 

Probability of occurrence of an event 

  = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

A leap has 366 days i.e. 52 weeks + 2 days. So, there will be 52 Sundays for sure (because every week has one Sunday) 

So, we want another Sunday from the remaining two days.

The two days may be Sunday, Monday or Monday, Tuesday or Tuesday, Wednesday or Wednesday, Thursday or Thursday, Friday or Friday, Saturday or Saturday, Sunday 

So, total outcomes are 7 and desired the outcomes are 2(Sunday, Monday or Saturday, Sunday) 

Therefore, the probability of getting 53 Sundays in a leap year 

\(\frac{2}{7}\)

Conclusion: Probability of getting 53 Sundays in a leap year is  \(\frac{2}{7}\)

We know that, 

Probability of occurrence of an event 

  = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

An ordinary year has 365 days i.e. it has 52 weeks + 1 day. So, there will be 52 Tuesdays for sure(because every week has 1 Tuesday) 

So, we want another Tuesday that to from that 1 day left(as there is only one Tuesday left after 52 weeks) 

This one day can be, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday. Of these total 7 outcomes, the desired outcome is 1, i.e. Tuesday 

Therefore, the probability of getting 52 Tuesdays in an ordinary year =  \(\frac{1}{7}\)

Conclusion: Probability of getting 53 Tuesdays in an ordinary year is \(\frac{1}{7}\)

We know that, 

Probability of occurrence of an event

  = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

Total no. of outcomes are 12 

Desired output is picking a number which is multiple of 2 or 3. So, desire outputs are 2, 3, 4, 6, 8, 9, 10, 12. Total no. of desired outputs are 8

Therefore, the probability of getting a number which is multiple of 2 or 3

= \(\frac{8}{12}\)

=   \(\frac{2}{3}\)

Conclusion: Probability of picking a ticket which is multiple of 2 or 3 is \(\frac{2}{3}\)

We know that, 

Probability of occurrence of an event

  = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

Total no. of outcomes are 52 

Desired output is a number greater than 3 and less than 10. 

There will be four sets of each card naming A, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. So, there will be a total of 24 cards between 3 and 10 

Therefore, the probability of picking card between 3 and 10 = \(\frac{24}{52}\) = \(\frac{6}{13}\)

Conclusion: Probability of picking a card between 3 and 10 is \(\frac{6}{13}\)

Total number of possible outcomes, n(S) = 50 

Number of events of getting a prime number, N(E) = 15 

Probability, 

P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{15}{50}\) = \(\frac{3}{10}\)

Given: A card is drawn at random from a pack of 52 cards

Required to Find: Probability of the following

Total number of cards in a pack = 52

(i) Number of cards which are black king = 2

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a black king = 2/52 = 1/26

(ii) Total number of black cards is (13 + 13) = 26

Total number of kings are 4 in which 2 black kings are also included.

So, the total number of black cards or king will be 26+2 = 28

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a black cards or a king = 28/52 = 7/13

(iii) Total number of cards which are black and a king cards is 2

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a black cards and a king is 2/52 = 1/26

(iv) A jack, queen or a king are 3 from each 4 suits.

So, the total number of a jack, queen and king are 12.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a jack, queen or a king is 12/52 = 3/13

(v) Total number of heart cards are 13 and king are 4 in which king of heart is also included.

So, the total number of cards that are a heart and a king = 13 + 3 = 16

Hence, the total number of cards that are neither a heart nor a king = 52 – 16 = 36

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting cards neither a heart nor a king = 36/52 = 9/13

(vi) Total number of spade cards is 13

Total number of aces are 4 in which ace of spade is included in the number of spade cards.

Hence, the total number of card which are spade or ace = 13 + 3 = 16

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting cards that is spade or an ace = 16/52 = 4/13

(vii) Total number of ace cards are 4 and king are 4

Total number of cards that are an ace or a king = 4 + 4 = 8

So, the total number of cards that are neither an ace nor a king is 52 – 8 = 44

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting cards which are neither an ace nor a king = 44/52 = 11/13

(viii) It’s know that the total number of red cards is 26.

Total number of queens are 4 in which 2 red queens are also included

Hence, total number of red cards or queen will be 26 + 2 = 28

So, the total number of cards that are neither a red nor a queen= 52 -28 = 24

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting neither a red card nor a queen = 24/52 = 6/13

(ix) Total number of card other than ace is 52 – 4 = 48

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting other than ace = 48/52 = 12/13

(x) Total number of tens in the pack of cards is 4.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a ten = 4/52 = 1/13

(xi) Total number of spade is 13

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a spade = 13/52 = 1/4

(xii) Total number of black cards in the pack is 26

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting black cards is 26/52 = 1/2

(xiii) Total number of 7 of club is 1 only.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a 7 of club = 1/52

(xiv) Total number of jacks are 4

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a jack = 4/52 = 1/13

(xv) Total number of ace of spade is 1

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting an ace of spade = 1/52

(xvi) Total number of queens is 4

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a queen = 4/52 = 1/13

(xvii) Total number of heart cards is 13

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a heart card = 13/52 = 1/4

(xviii) Total number of red cards is 26

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a red card = 26/52 = 1/2

We know that,

Probability of occurrence of an event

 = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

By permutation and combination, total no. of ways to pick r objects from given n objects is ⁿC_r 

Now, total no. of ways to pick a ball from 20 balls is 20_c1 = 20 

Our desired output is to pick a white or red ball. So, no. of ways to pick a white or red ball from 16 balls(because there are a total of 16 balls which are either red or white) is ¹⁶c₁ = 16 

Therefore, the probability of picking a white or red ball = \(\frac{16}{20}\) = \(\frac{4}{5}\) 

Conclusion: Probability of picking a white or red ball from 9 red, 7 white, and 4 black balls is \(\frac{4}{5}\)

We know that, 

Probability of occurrence of an event 

 = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

By permutation and combination, total no. of ways to pick r objects from given n objects is ⁿC_r 

Now, total no. of ways to pick a ball from 20 balls is ²⁰c₁ = 20

Our desired output is to pick a white or red ball. So, no. of ways to pick a white or red ball from 16 balls(because there are a total of 16 balls which are either red or white) is ¹⁶c₁ = 16 

Therefore, the probability of picking a white or black ball = \(\frac{11}{20}\) 

Conclusion: Probability of picking a white or black ball from 9 red, 7 white, and 4 black balls is  \(\frac{11}{20}\) 

Given: Tickets are marked numbers from 1 to 50. And, one ticket is drawn at random.

Required to find: Probability of getting a prime number on the drawn ticket

Total number of tickets is 50.

Tickets which are number as prime number are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

Total number of tickets marked as prime is 15.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a prime number on the ticket = 15/50 = 3/10

We know that, 

Probability of occurrence of an event

 = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

By permutation and combination, total no. of ways to pick r objects from given n objects is ⁿC_r 

Now, total no. of ways to pick a ball from 20 balls is ²⁰c₁ = 20 

Our desired output is to pick a white ball. So, no. of ways to pick a white ball from 7 white balls(because the white ball can be picked from only white balls) is ⁷c₁ = 7 

Therefore, the probability of picking a white ball = \(\frac{7}{20}\)

Conclusion: Probability of picking a white ball from 9 red, 

7 white and 4 black balls is \(\frac{7}{20}\)

Sample space, n(S) = 15 

Number of events of getting numbers multiple of 4, 

n(E) = 3

∴ P(E) = \(\frac{n(E)}{n(S)} = \frac{3}{15}\) = \(\frac{1}5\)

Given: A bag contains 10 red and 8 white balls

Required to find: Probability that one ball is drawn at random and getting a white ball

Total number of balls 10 + 8 = 18

Total number of white balls is 8

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a white ball from the urn is 8/18 = 4/9

Given: Numbers are from 1 to 15. One number is selected

Required to find: Probability that the selected number is a multiple of 4

Total number between from 1 to 15 to 15

Numbers that are multiple of 4 are 4, 8 and 12.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of selecting a number which a multiple of 4 is 3/15 = 1/5

Given: A bag contains 6 red, 8 black and 4 white balls and a ball is drawn at random

Required to find: Probability that the ball drawn is not black

Total number of balls 6 + 8 + 4 = 18

Total number of black balls is 8

So, the total number of balls which are not black is 18 – 8 = 10

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a ball which is not black = 10/18 = 5/9

Sample space, n(S) = 18 

Number of events of getting balls not black, 

n(E) = 10

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{10}{18}\) = \(\frac{5}9\)

Given: A bag contains 7 red and 5 white balls and a ball is drawn at random

Required to find: Probability that the ball drawn is white

Total number of balls 7 + 5 = 12

Total number of white balls is 5

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a white ball = 5/12

Sample space, n(S) = 12 

Number of events of getting white ball, 

n(E) = 5

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{5}{12}\)

Here, n(S) = 15 

(i) n(E) = 6

∴P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{6}{15}\) = \(\frac{2}{5}\)

(ii) n(E) = 4

∴P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{15}\)

(iii) number of events of getting black balls, 

n(E) = 5 

∴ probability of getting black balls, P(E) =   \(\frac{n(E)}{n(S)}\) = \(\frac{5}{15}\) = \(\frac{1}{3}\)

Hence, 

probability of getting black balls = 1 – P(E) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)

(iv) number of events of getting a red or white ball, 

n(E) = 10

 ∴P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{10}{15}\) = \(\frac{2}{3}\)

Sample space, n(S) = 20 

Number of events of getting a multiples of 3 or 7 on the drawn ticket,

n(E) = 8 {3,6,9,12,15,18,7,14}

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{8}{20}\) = \(\frac{2}5\)

Sample space, n(S) = 15 

(i) n(E) = 7

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{7}{15}\)

(ii) n(E) = 10

 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{8}{15}\)

(iii) n(E) = 5

 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{5}{15}\) = \(\frac{1}{3}\)

Sample space, n(S) = 35 

Number of events of getting a prize, 

n(E) = 10

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{10}{35}\) = \(\frac{2}{7}\)

We not that, 

P(E) + P’(E) = 1 

Here, 

P(E) = 0.3 

P’(E) = 1 – 0.3 = 0.7

Correct option is (C) Certain event

If the probability of an event is 1, then the event is called the certain event (or sure event).

(C) Certain event

P(A) = 0.72 

P(A’) = 1 – 0.72 

= 0.28

Correct option is (B) 1/2

Possible outcomes of experiment of tossing a coin are S = {H, T}.

\(\therefore\) n (S) = 2

n (H) = 1, n (T) = 1

\(\therefore\) Probability of getting head on upper side is \(P(H)=\frac{n(H)}{n(S)}=\frac12.\)

Correct option is  B) 1/2