RS Aggarwal Solutions Class 10 Ex 15C

QUESTION 1:A bag contains 18 balls out of which x balls are red.

(i) If one ball is drawn at random from the bag, what is the probability that it is not red? 

(ii) If two more red balls are put in the bag, the probability of drawing a red ball will be \(\frac{9}{8}\) times the probability of drawing a red ball in the first case. Find the value of x.

Solution:

The total number of balls = 18.

Number of red balls = x.

(i) Number of balls which are not red = 18 – x

Therefore,  P(getting a ball which is not red) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{18-x}{18}\)

Thus, the probability of drawing a ball which is not red is \(\frac{18-x}{18}\).

(ii) Now, total number of balls = 18 + 2 = 20.

Number of red balls now = x + 2.

P(getting a red ball now) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{x+2}{20}\)

And P(getting a red ball in first case) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{x}{18}\)

Since, it is given that probability of drawing a red ball now will be \(\frac{9}{8}\) times the probability of drawing a red ball in the first case.

Thus, \(\frac{x+2}{20}=\frac{9}{8}\times \frac{x}{18}\)

\(\Rightarrow\) 144(x + 2) = 180x

\(\Rightarrow\) 144x + 288 = 180x

\(\Rightarrow\) 36x = 288

\(\Rightarrow\) x = \(\frac{288}{36}\) = 8

Thus, the value of x is 8.

QUESTION 2:A jar contains 24 marbles. Some of these are green others are blue. If a marble is drawn at random from the jar, the probability that it is green is \(\frac{2}{3}\). Find the number of blue marbles in the jar.

Solution:

The total number of marbles = 24.

Let the number of blue marbles be x.

Then, the number of green marbles = 24 – x

Therefore, P(getting a green marble) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{24-x}{24}\)

But, P(getting a green marble) = \(\frac{2}{3}\)            [Given]

Therefore, \(\frac{24-x}{24}=\frac{2}{3}\)

\(\Rightarrow\) 3(24 – x) = 48

\(\Rightarrow\) 72 – 3x = 48

\(\Rightarrow\) 3x = 72 – 48

\(\Rightarrow\) 3x = 24

\(\Rightarrow\) x = 8

Thus, the number of blue marbles in the jar is 8.

QUESTION 3:A jar contains 54 marbles, each of which some are  blue, some are green and some are white. The probability of selecting a blue marble at random is \(\frac{1}{3}\) and the probability of selecting a green marble at random is \(\frac{4}{9}\). How many white marbles does the jar contain?

Solution:

The total number of marbles = 54.

It is given that,

P(getting a blue marble) = \(\frac{1}{3}\) and P(getting a green marble) = \(\frac{4}{9}\)

Let P(getting a white marble) be x.

Since, there are only 3 types of marbles in the jar, the sum of probabilities of all three marbles must be 1.

Therefore, \(\frac{1}{3}+\frac{4}{9}+x=1\)

\(\Rightarrow\) \(\frac{3+4}{9}+x=1\)

\(\Rightarrow\) x = \(1-\frac{7}{9}\)

\(\Rightarrow\) x = \(\frac{2}{9}\)

Therefore,  P(getting a white marble) = \(\frac{2}{9}\) . . . . . . . (1)

Let the number of white marbles be n.

Then, P(getting a white marbles) = \(\frac{n}{54}\) . . . . . . . (2)

From (1) and (2),

\(\frac{n}{54}\) = \(\frac{2}{9}\)

\(\Rightarrow\) n = \(\frac{2\times 54}{9}\)

\(\Rightarrow\) n = 12

Thus, there are 12 white marbles in the jar.

QUESTION 4:A carton consists of 100 shirts of which 88 are good and 8 have minor defects. Rohit, a trader, will only accept the shirts which are good. But, Kamal, and another trader will only reject the shirts which have major defects. 1 shirt is drawn at random from the carton. What is the probability that it is acceptable to

(i) Rohit,

(ii) Kamal?

Solution:

Total numbers of shirts = 100.

The number of good shirts = 88.

The number of shirts with minor defects = 8.

Number of shirts with major defects = 100 – 88 – 8 = 4.

(i) P(the drawn shirt is acceptable to Rohit) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{88}{100}\) = \(\frac{22}{25}\)

Thus, the probability that the drawn shirt is acceptable to Rohit is \(\frac{22}{25}\).

(ii) P(the drawn shirt is acceptable to Kamal) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{88+8}{100}\) = \(\frac{96}{100}\) = \(\frac{24}{25}\)

Thus, the probability that the dwarn shirt is acceptable to Kamal is \(\frac{24}{25}\).

QUESTION 5:A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is

(i) extremely patient,

(ii) extremely kind or honest.

Which of the above values did you prefer more?

Solution:

The total number of persons = 12.

The number of persons who are extremely patient = 3.

The number of persons who are extremely honest = 6.

Number of persons who are extremely kind = 12 – 3 – 6 = 3.

(i) P(selecting a person who is extremely patient) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)

Thus, the probability of selecting a person who is extremely patient is \(\frac{1}{4}\).

(ii) P(selecting a person who is extremely kind or honest) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{6+3}{12}\) = \(\frac{9}{12}\) = \(\frac{3}{4}\)

Thus, the probability of selecting a person who is extremely kind or honest is \(\frac{3}{4}\).

From the three given values, we prefer honesty more.

QUESTION 6:A die is rolled twice. Find the probability that

(i) 5 will not come up either time,

(i) 5 will come up exactly one time,

(iii) 5 will come up both the times.

Solution:

Total number of outcomes = 36.

(i) Cases where 5 comes up on at least one time are (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) and (6, 5).

The number of such cases = 11.

The number of cases where 5 will not come up either time = 36 – 11 = 25.

Therefore,  P(5 will not come up either time) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{25}{36}\)

Thus, the probability that 5 will not come up either time is \(\frac{25}{36}\).

(ii) Cases where 5 comes up on exactly one time are (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6) and (6, 5).

The number of such cases = 10.

Therefore,  P(5 will come up exactly one time) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{10}{36}\) = \(\frac{5}{18}\)

Thus, the probability that 5 will come up exactly one time is \(\frac{5}{18}\).

(iii) Cases, where 5 comes up on exactly two times, is (5, 5).

The number of such cases = 1.

Therefore, P(5 will come up both the times) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{1}{36}\)

Thus, the probability that 5 will come up both the times is \(\frac{1}{36}\).

QUESTION 7:Two dice are rolled once. Find the probability of getting such numbers on 2 dice whose product is a perfect square.

Solution:

Number of possible outcomes = 36

Let E be the event of getting two numbers whose product is a perfect square.

Then, the favorable outcomes are (1, 1), (1, 4), (4, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

The number of favorable outcomes = 8.

Therefore, P(getting numbers whose product is a perfect square) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{8}{36}\) = \(\frac{2}{9}\)

Thus, the probability of getting such numbers on two dice whose product is a perfect square is \(\frac{2}{9}\).

QUESTION 8:A letter is chosen at random from the letter of the word ‘ASSOCIATION’. Find the probability that the chosen letter is a

(i) vowel

(ii) consonant

(iii)  S

Solution:

Total numbers of letters in the given word ASSOCIATION = 11

(i) Number of vowels (A, O, I, A, I, O) in the given word = 6

Therefore, P(getting a vowel) = \(\frac{6}{11}\)

(ii) Number of consonants in the given word (S, S, C, T, N) = 5

Therefore, P(getting a consonant) = \(\frac{5}{11}\)

 

(iii) Number of S in the given word = 2

Therefore, P(getting an S) = \(\frac{2}{11}\)

QUESTION 9:5 cards —— the ten, jack, queen, king and ace of diamonds are well shuffled with their faces downward. One card is then picked up at random.

(a) What is the probability that the drawn card is the queen?

(b) If the queen is drawn and put aside and a second card is drawn, find the probability that the second card is

(i) an ace,

(ii) a queen.

Solution:

Total number of card = 5.

(a) Number of queens = 1.

Therefore, P(getting a queen) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{1}{5}\)

Thus, the probability that the drawn card is the queen is \(\frac{1}{5}\).

 

(b) When the queen has put aside, number of remaining cards = 4.

(i) The number of aces = 1.

Therefore,  P(getting an ace) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{1}{4}\)

Thus, the probability that the drawn card is an ace is \(\frac{1}{4}\).

(ii) Number of queens = 0.

Therefore, P(getting a queen now) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{0}{4}\) = 0

Thus, the probability that the drawn card is a queen is 0.

QUESTION 10:A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that the card was drawn is neither a red card nor a queen.

Solution:

Total number of all possible outcomes = 52

There are 26 red cards (including 2 queens) and apart from these, there are 2 more queens.

Number of cards, each one of which is either a red card or a queen = 26 + 2 = 28

Let E be the event that the card drawn is neither a red card nor a queen.

Then, the number of favorable outcomes = (52 – 28) = 24

Therefore, P(getting neither a red card nor a queen) = P(E) = \(\frac{24}{52}\) = \(\frac{6}{13}\)

QUESTION 11:What is the probability that an ordinary year has 53 Mondays?

Solution:

An ordinary year has 365 days consisting of 52 weeks and 1 day.

This day can be any day of the week.

Therefore, P(of this day to be Monday) = \(\frac{1}{7}\)

Thus, the probability that an ordinary year has 53 Mondays is \(\frac{1}{7}\).

QUESTION 12:All red face cards are removed from a pack of playing cards. The remaining cards are well-shuffled and then a card is drawn at random from them. Find the probability that the drawn card is

(i) a red card,

(ii) a face card,

(iii) a card of clubs.

Solution:

There are 6 red face cards. These are removed.

Thus, remaining number of card = 52 – 6 = 46.

(i) Number of red cards now = 26 – 6 = 20.

Therefore,  P(getting a red card) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{20}{46}\) = \(\frac{10}{23}\).

Thus, the probability that the drawn card is a red card is \(\frac{10}{23}\).

(ii) Number of face cards now = 12 – 6 = 6.

Therefore,  P(getting a face card) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{6}{46}\) = \(\frac{3}{23}\).

Thus, the probability that the drawn card is a face card is \(\frac{3}{23}\).

(iii) The number of card of clubs = 12.

Therefore,  P(getting a card of clubs) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{12}{46}\) = \(\frac{6}{23}\)

Thus, the probability that the drawn card is a card of clubs is \(\frac{6}{23}\).

QUESTION 13:All kings, queens, and aces are removed from a pack of 52 cards. The remaining cards are well-shuffled and then a card is drawn from it. Find the probability that the drawn card is

(i) a black face card,

(ii) a red face card.

Solution:

These are 4 kings, 4 queens, and 4 aces. These are removed.

Thus, remaining number of cards = 52 – 4 – 4 – 4 = 40.

(i) Number of black face cards now = 2 (only black jacks).

Therefore,  P(getting a black face card) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{2}{40}\) = \(\frac{1}{20}\)

Thus, the probability that the drawn card is a black face card is  \(\frac{1}{20}\).

(ii) Number of red cards now = 26 – 6 = 20.

Therefore,  P(getting a red card) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{20}{40}\) = \(\frac{1}{2}\)

Thus, the probability that the drawn card is a red card is \(\frac{1}{2}\).

QUESTION 14:A game consists of tossing a 1 rupee coin three times, and noting its outcomes each time. Find the probability of getting

(i) 3 heads,

(ii) at least 2 tails.

Solution:

When a coin is tossed three times, all possible outcomes are

HHH, HHT, HTH, THH, HTT, THT, TTH and TTT.

The number of total outcomes = 8.

(i) The outcome with three heads is HHH.

The number of outcomes with three heads = 1.

Therefore,  P(getting three heads) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{1}{8}\)

Thus, the probability of getting three heads is \(\frac{1}{8}\).

(ii) Outcomes with at least two tails are TTH, THT, HTT and TTT.

The number of outcomes with at least two tails = 4.

Therefore,  P(getting at least two tails) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{4}{8}\) = \(\frac{1}{2}\)

Thus, the probability of getting at least two tails is \(\frac{1}{2}\).

QUESTION 15:Find the probability that a leap year selected at random will contain 53 Sundays.

Solution:

A leap year has 366 days with 52 weeks and 2 days.

Now, 52 weeks contains 52 Sundays.

The remaining two days can be:

(i) Sunday and Monday

(ii) Monday and Tuesday

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(vi) Friday and Saturday

(vii) Saturday and Sunday

Out of these 7 cases, there are two cases favoring it to be Sunday.

Therefore, P(a leap year having 53 Sundays) = \(\frac{Number\, of\, favorable\, outcomes}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{2}{7}\)

Thus, the probability that a leap year selected at random will contain 53 Sundays is \(\frac{2}{7}\).


Practise This Question

Find the value of k for which x24x+k=0 has coincident roots.