Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 13 Probability InText Questions to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 13 Probability InText Questions

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(a) Outcomes of which of the following experiments cure equally likely. (AS_{3})(Page No. 307)

Question 1.

Getting a digit 1, 2, 3, 4, 5 or 6 when a die is rolled.

Solution:

Equally likely

Question 2.

Selecting a different colour ball from a bag of 5 red balls, 4 blue balls and 1 black ball.

Note : Picking two different colour balls, i.e., Picking a red, a blue (or) black ball from a ………

Solution:

Not equally likely

Question 3.

Winning in a game of carrom.

Solution:

Not equally likely

Question 4.

Units place of a two digit number selected may be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9.

Solution:

equally likely

Question 5.

Selecting a different colour ball from a bag of 10 red balls, 10 blue balls and 10 black balls.

Solution:

equally likely

Question 6.

Raining on a particular day of July.

Solution:

equally likely

(b) Are the outcomes of every experiment equally likely ?

Solution:

Outcomes of all experiments need not necessarily be equally likely.

(c) Give examples of 5 experiments that have equally likely outcomes and five more examples that do not have equally likely outcomes.

Solution:

Equally likely events :

a) Getting an even or odd number when a die is rolled.

b) Getting tail or head when a coin is tossed.

c) Getting an even (or) odd number when a card is drawn at random from a pack of cards numbered from 1 to 10.

d) Picking a green or black ball from a bag containing 8 green balls and 8 black balls.

e) Selecting a boy or girl from a class of 20 boys and 20 girls.

f) Selecting a red or black card from a deck of cards.

Events which are not equally likely :

a) Getting a prime (or) composite number when a die is thrown.

b) Getting an even or odd number when a card is drawn at random from a pack of cards numbered from 1 to 5.

c) Getting a number which is a multiple of 3 (or) not a multiple of 3 from numbers 1, 2, …….., 10.

d) Getting a number less than 5 or greater than 5.

e) Drawing a white ball or green ball from a bag containing 5 green balls and 8 white balls.

Think of 5 situations with equally likely events and find the sample space.

a) Tossing a coin : Getting a tail or head when a coin is tossed.

Sample space = {T, H}

b) Getting an even (or) odd number when a die is rolled.

Sample space = {1, 2, 3, 4, 5, 6}

c) Winning a game of shuttle Sample space = {win, loss}

d) Picking a black (or) blue ball from a bag containing 3 blue balls and 3 black balls = {blue, black}

e) Drawing a red coloured card or black coloured card from a deck of cards = {black, red}

d) Is getting a head complementary to getting a tail ? Give reasons.

Solution:

Number outcomes favourable to head = 1

Probability of getting a head = \(\frac{1}{2}\) [P(E)]

Number of outcomes not favourable to head = 1

Probability of not getting a head = \(\frac{1}{2}\) (P (\(\overline{\mathrm{E}}\)))

Now P(E) + P(\(\overline{\mathrm{E}}\)) = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1

∴ Getting a head is complementary to getting a tail.

e) In case of a die is getting a 1 complementary to events getting 2, 3, 4, 5, 6 ? Give reasons for your answer.

Solution:

Yes, complementary events

∴ Probability of getting a 1 = \(\frac{1}{6}\) [P(E)]

Probability of getting 2, 3, 4, 5, 6

= P (\(\overline{\mathrm{E}}\)) = \(\frac{5}{6}\)

p(E) + P(\(\overline{\mathrm{E}}\)) = \(\frac{1}{6}\) + \(\frac{5}{6}\) = \(\frac{6}{6}\) = 1

f) Write of five new pair of events that are complementary.

Solution:

a) When a die is thrown, getting an even number is complementary to getting an odd number.

b) Drawing a red card from a deck of cards is complementary to getting a black card.

c) Getting an even number is complementary to getting an odd number from numbers 1, 2, ……….., 8.

d) Getting a Sunday is complementary to getting any day other than Sunday in a week.

e) Winning a running race is complementary to loosing it.

Try This

Question 1.

A child has a die whose six faces show the letters A, B, C, D, E and F. The die is thrown once. What is the probability of getting (i) A ? (ii) D ? (AS_{4})(Page No. 312)

Solution:

Total number of outcomes [A, B, C, D, E and F] = 6

i) Number of favourable outcomes to A = 1

Probability of getting A = P(A)

= \(\frac{1}{6}\)

ii) Number of outcomes favourable to D = 1

Probability of getting D

= P(D)

= \(\frac{1}{6}\)

Question 2.

Which of the following cannot be the probability of an event ? (AS_{3})(Page No. 312)

a) 2.3

b) – 1.5

c) 15%

d) 0.7

Solution:

a) 2.3 – Not possible

b) -1.5 – Not possible

c) 15% – May be the probability

d) 0.7 – May be the probability

Question 3.

You have a single deck of well shuffled cards. Then,

i) What is the probability that the card drawn will be a queen ? (AS_{4})(Page No. 313)

Solution:

Number of all possible outcomes

= 4 × 13 = 1 × 52 = 52

Number of outcomes favourable to Queen

= 4[♥Q ♥Q ♥Q ♥Q]

Probability P(E)

= \(\frac{\text { No. of favourable outcomes }}{\text { Total no.of outcomes }}\)

= \(\frac{4}{52}\)

= \(\frac{1}{13}\)

ii) What is the probability that it is a face card ? (Page No. 314)

Solution:

Face cards are J, Q, K.

∴ Number of outcomes favourable to face cards = 4 × 3 = 12

No. of all possible outcomes = 52

P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)

= \(\frac{12}{52}\) = \(\frac{3}{13}\)

iii) What is the probability that it is a spade ? (Page No. 314)

Solution:

Number of spade cards = 13

Total number of cards = 52

Probability = \(\frac{\text { Number of outcomes favourable to spade }}{\text { Number of all outcomes }}\)

= \(\frac{13}{52}\) = \(\frac{1}{4}\)

iv) What is the probability that is the face cards of spades ? (Page No. 314)

Solution:

Number of outcomes favourable to face cards of spades = (K, Q, J) = 3

Number of all outcomes = 52 3

∴ P(E) = \(\frac{3}{52}\)

v) What is the probability it is not a face card ? (Page No. 314)

Solution:

Probability of a face card = \(\frac{12}{52}\)

∴ Probability that the card is not a face card

= 1 – \(\frac{12}{52}\) [P (\(\overline{\mathrm{E}}\)) = 1 – P(E)]

= \(\frac{52-12}{52}\)

= \(\frac{40}{52}\) = \(\frac{10}{13}\)

(Or)

Number of favourable outcomes = 4 × 10 = 40

Number of all outcomes = 52

∴ Probability = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)

= \(\frac{40}{52}\) = \(\frac{10}{13}\)

Think – Discuss

Question 1.

Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of any game ? (Page No. 312)

Solution:

Probability of getting a head is \(\frac{1}{2}\) and a tail is

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

Hence, tossing a coin is a fair way.

Question 2.

Can \(\frac{7}{2}\) be the probability of an event ? Explain. (AS_{3}) (Page No. 312)

Solution:

\(\frac{7}{2}\) can’t be the probability of any event. Since probability of any event should lie between 0 and 1.

Question 3.

Which of the following arguments are correct and which are not correct ? Given reasons. (Page No. 312)

i) If two coins are tossed simultaneously, there are three possible outcomes – two heads, two tails (or) one of each. Therefore, for each. If these outcomes, the probability is \(\frac{1}{3}\)

Solution:

False

Reason :

All possible outcomes are 4. They are HH, HT, TH, TT

Thus, probability of two heads = \(\frac{1}{4}\)

Probability of two tails = \(\frac{1}{4}\)

Probability of one each = \(\frac{2}{4}\) = \(\frac{1}{2}\)

ii) If a die is thrown, there are two possible outcomes – an odd number (or) an even number. Therefore, the probability of getting an odd number is \(\frac{1}{2}\).

Solution:

True

Reason :

All possible outcomes = (1, 2, 3, 4, 5, 6) = 6

Outcomes favourable to an odd number = (1, 3, 5) = 3

Outcomes favourable to an even number = (2, 4, 6) = 3

∴ Probability (odd number)

= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)

= \(\frac{3}{6}\) = \(\frac{1}{2}\)