AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 13 Probability Ex 13.1 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability Exercise 13.1

### 10th Class Maths 13th Lesson Probability Ex 13.1 Textbook Questions and Answers

Question 1.

Complete the following statements:

i) Probability of an event E + Probability of the event ‘not E’ = 1 .

ii) The probability of an event that cannot happen is zero.

Such an event is called an impossible event.

iii) The probability of an event that is certain to happen is 1 such an event is called sure or certain event.

iv) The sum of the probabilities of all the elementary events of an experiment is 1 .

v) The probability of an event is greater than or equal to zero and less than or equal to 1 .

Question 2.

Which of the following experiments have equally likely outcomes? Explain.

i) A driver attempts to start a car. The car starts or does not start.

Answer:

Equally likely. Since both have the same probability \(\frac{1}{2}\).

ii) A player attempts to shoot a basket-ball. She/he shoots or misses the shot.

Answer:

Equally likely. Since both have the same probability \(\frac{1}{2}\).

iii) A trial is made to answer a true-false question. The answer is right or wrong.

Equally likely. Since both have the same probability \(\frac{1}{2}\).

iv) A baby is born. It is a boy or a girl.

Equally likely. Since both the events have the same probability \(\frac{1}{2}\).

Question 3.

If P(E) = 0.05, what is the probability of not E?

Answer:

Given: P(E) = 0.05

Hence, P(E) + P(\(\overline{\mathrm{E}}\)) = 1, where P(\(\overline{\mathrm{E}}\)) is the probability of ‘not E’

0.05 + P(\(\overline{\mathrm{E}}\)) = 1

∴ P(\(\overline{\mathrm{E}}\)) = 1 -0.05 = 0.95.

Question 4.

A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

i) an orange flavoured candy?

ii) a lemon flavoured candy?

Answer:

Bag contains only lemon flavoured candies.

i) Taking an orange flavoured candy is an impossible event and hence the probability is zero.

ii) Also taking a lemon flavoured candy is a sure event and hence its probability is 1.

Question 5.

Rahim removes all the hearts from the cards. What is the probability of

i. Picking out an ace from the remaining pack.

ii. Picking out a diamond.

iii. Picking out a card that is not a heart.

iv. Picking out the Ace of hearts.

Answer:

Total number of cards in the deck = 52.

Total number of hearts in the deck of cards =13.

When Hearts are removed, remaining cards = 52 – 13 = 39.

i)Picking out an Ace:

Number of outcomes favourable to Ace = 3 [∵ ♦ A, ♥ A, ♠ A, ♣ A]

Total number of possible outcomes from the remaining cards = 39

– after removing Hearts.

Probability = P(A)

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

= \(\frac{3}{39}\) = \(\frac{1}{13}\)

ii) Picking out a diamond:

Number of favourable outcomes to diamonds (♦) = 13

Total number of possible outcomes = 39

∴ p(♦) = \(\frac{13}{39}\) = \(\frac{1}{3}\)

iii) Picking out a card that is ‘not a heart’:

As all hearts are removed, the remain-ing cards are all non-heart cards. So the picked card will be definitely a non-heart card. So this is a sure event.

Hence its probability is one

P(E) = \(\frac{39}{39}\) = 1

iv) Picking out the Ace of Hearts:

a) As all the heart cards are removed the left over cards will have three suits (i) spades, (ii), clubs, (iii) dia¬monds of each 13.

Hence total outcomes = 3 × 13 = 39 But among them there is no Ace of heart. So number of favourable outcomes for picking Ace of heart = zero.

∴ Probability P(E) = \(\frac{0}{39}\) = 0

So it is an impossible event.

b) If picking from the rest of the cards, it is an impossible event and hence probability is zero.

It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992.

Question 6.

What is the probability that the 2 students have the same birthday?

Answer:

Let P(E) = The probability that two students not having the same birthday = 0.992

Then P(\(\overline{\mathrm{E}}\)) = The complementary event of E, i.e., two students having the same birthday Also, P(E) + p(\(\overline{\mathrm{E}}\)) = 1

∴ The probability that two students have the same birthday P(\(\overline{\mathrm{E}}\)) = 1 – P(E)

= 1 – 0.992 = 0.008

Question 7.

A die is thrown once. Find the probability of getting

(i) a prime number;

(ii) a number lying between 2 and 6;

(iii) an odd number.

Answer:

i) When a die is thrown for one time, total number of outcomes = 6

No. of outcomes favourable to a prime number (2, 3, 5) = 3

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

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

ii) No. of outcomes favourable to a number lying between 2 and 6 (3, 4, 5) = 3

∴ Probability of getting a number between 2 and 6

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

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

iii) Number of outcomes favourable to an odd number (1, 3, 5) = 3

∴ Probability of getting an odd number P(odd)

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

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

Question 8.

What is the probability of drawing out a red king from a deck of cards?

Answer: Number of favourable outcomes to red king (♥ K, ♦ K) = 2.

Number of total outcomes = 52

(∵ Number of cards in a deck of cards = 52)

∴ Probability of getting a red king P (Red king)

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

= \(\frac{2}{52}\) = \(\frac{1}{26}\)

Question 9.

Make 5 more problems getting probability using dice, cards or birthdays and discuss with friends and teacher about their solutions.

Answer:

Class-room activity.