The document Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.

All you need of Class 10 at this link: Class 10

**Q1. Complete the following statements.****(i) Probability of an event E + Probability of the event ‘not E’ = ________(ii) The probability of an event that cannot happen is ________. Such an event is called ________.(iii) The probability of an event that is certain to happen is ________ Such an event is called ________.(iv) The sum of the probabilities of all the elementary events of an experiment is ________.(v) The probability of an event is greater than or equal to ________ and less than or equal to ________.**

**Q2. 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.(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.(iii) A trial is made to answer a true-false question. The answer is right or wrong.(iv) A baby is born. It is a boy or a girl.**

**Q3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?****Ans: **Since on tossing a coin, the outcomes ‘head’ and ‘tail’ are equally likely, the result of tossing a coin is completely unpredictable and so it is a fairway.

**Q4. Which of the following cannot be the probability of an event?****(a) 2/3 ****(b) − 1.5 ****(c) 15% ****(d) 0.7**

**Ans: **(b)**Solution:** Since the probability of an event cannot be negative.**∴** Option **(b)**** −1.5** cannot be the probability of an event.

**Q5. If P(E) = 0.05, what is the probability of ‘not E’?****Ans:** ∵ P(E) + P(not E)=1

∴ 0.05 + P(not E)=1 ⇒ P(not E) = 1 − 0.05

= 0.95

Thus, the probability of ‘not E’ = 0.95.

**Q6. 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?**

∴ Taking out any orange flavoured candy is not possible.

⇒ Probability of taking out an orange flavoured candy = 0.

**Q7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?****Ans:** ∴ Let the probability of 2 students having same birthday = P(SB) And the probability of 2 students not having the same birthday = P(nSB)

∴ P(SB) + P(nSB)=1

⇒ P(SB) + 0.992 = 1

⇒ P(SB)=1 − 0.992 = 0.008

So, the required probability of 2 boys having the same birthday = **0.008.**

**Q8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?****Ans:**

Total number of balls = 3 + 5 = 8

∴ Number of all possible outcomes = 8**(i) For red balls**

There are 3 red balls.

∴ Number of favourable outcomes = 3**(ii) For not red balls**

**Q9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? and (ii) white? (iii) not green?****Ans:**

Total number of marbles = 5 + 8 + 4 = 17**(i) For red marbles**

∵ Number of red marbles = 5

∴ Number of favourable outcomes = 5

∴ Probability of red marbles, P(red) = 5/17**(ii) For white balls**

∵ Number of white balls = 8

∴ Probability of white balls,

∴ P(white)= 8/17**(iii) For not green balls **

∵ Number of white balls = 4

∴ Number of ‘not green’ balls = 17 − 4 = 13

i.e., Favourable outcomes = 13

P(not green) = 13/17

OR

Number of green marbles = 4

∴ Number of ‘not green balls’ = 17 − 4 = 13

⇒ Favourable outcomes = 13

∴ P(not green) = 13/17

**Q10. A piggy bank contains hundred 50p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50p coin? and (ii) will not be Rs 5 coin?****Ans: **

Number of:

50 p coins = 100

Re 1 coins = 50

Rs 2 coins = 20

Rs 5 coins = 10

Total number of coins = 100 + 50 + 20 + 5 = 180**(i) For a 50 p coin:**Favourable events = 100

∴ P(50 p) = 100/180 = 5/9

∴ Number of ‘not Rs 5’ coins = 180 − 10 = 170

⇒ Favourable outcomes = 170

∴ P(not 5 rupee coin) = 170/180 = 17/18.

**Q11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig.). What is the probability that the fish taken out is a male fish?**

**Ans: **

Number of:

Male fishes = 5

Female fishes = 8

∴ Total number of fishes = 5 + 8 = 13

⇒ Total number of outcomes = 13**For a male fish:**Number of favourable outcomes = 5

∴ P(male fish) = 5/13.

**Q12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see figure), and these are equally likely outcomes. What is the probability that it will point at ****(i) 8?(ii) an odd number?(iii) a number greater than 2?(iv) a number less than 9?**

Total number of outcomes = 8

Number of favourable outcomes = 1

∴ P(8)

= 1/8

**(ii) When pointer points at an odd number**

Number of odd numbers from 1 to 8 = 4

[∵ Odd numbers are 1, 3, 5 and 7]

⇒ Number of favourable outcomes = 4

∴ P(odd) =

**(iii) When pointer points at a number greater than 2 **

Number of numbers greater than 2 = 6

[∴ The numbers 2, 3, 4, 5, 6, 7 and 8 are greater than 2]

⇒ Number of favourable outcomes = 6

∴ P(greater than 2) =

**(iv) When pointer points a number less than 9: **

Number of numbers less than 9 = 8

[a The numbers 1, 2, 3, 4, 5, 6, 7 and 8 are less than 9]

∴ Number of favourable outcome = 8

P(greater than 9)

**Q13. 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.**

∴ Number of total outcomes = 6

Since 2, 3, and 5 are prime number,

∴ Favourable outcomes = 3

P(prime) =

**(ii) For a number lying between 2 and 6 **

Since the numbers between 2 and 6 are 3, 4 and 5

∴ Favourable outcomes = 3**(iii) For an odd number**

Since 1, 3 and 5 are odd numbers.

⇒ Favourable outcomes = 3

**Q14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting(i) a king of red colour(ii) a face card(iii) a red face card(iv) the jack of hearts(v) a spade(vi) the queen of diamonds **

∴ Total number of possible outcomes = 52

∵ Number of red colour kings = 2

[∵ Kings of diamond and heart are red]

∴ Number of favourable outcomes = 2

E(red king)

**(ii) For a face card **

∵ 4 kings, 4 queens and 4 jacks are face cards

∴ Number of face cards = 12

⇒ Number of favourable outcomes = 12

∴ P(face)

** **

**(iii) For a red face card**

Since cards of diamond and heart are red

∴ There are [2 kings, 2 queens, 2 jacks] 6 cards are red

⇒ Favourable outcomes = 6

∴ P(red face)

**(iv) For a jack of hearts **

Since there is only 1 jack of hearts.

∴ Number of favourable outcomes = 1

P(jack of hearts)

**(v) For a spade**

∵There are 13 spades in a pack of 52 cards:

∴ Favourable outcomes are 13.

P(spade)

**(vi) For the queen of diamonds**

∵ There is only one queen of a diamond.

∴ Number of favourable outcomes.

P(queen of diamonds)

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

132 docs