CLASS 10th, PROBABILITY (NCERT)
Exercise: 15.1 (Solution 1), Question-1 to Question-12
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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 __________.
Solution:-
(i) Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) The probability of an event that cannot happen is 0. 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 a 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 0 and less than or equal to 1.
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 Solution: a true-false question. The Solution: is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:-
(i) This statement does not have equally likely outcomes as the car may or may not start depending upon various factors like fuel, etc.
(ii) Even this statement does not have equally likely outcomes as the player may shoot or miss the shot.
(iii) This statement has equally likely outcomes as it is known that the solution is either right or wrong.
(iv) This statement also has equally likely outcomes as it is known that the newly born baby can either be 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?
Solution:-
Tossing of a coin is a fair way of deciding because the number of possible outcomes are only 2 i.e. either head or tail. Since these two outcomes are an equally likely outcome, tossing is unpredictable and is considered to be completely unbiased.
Q4. Which of the following cannot be the probability of an event?
(A) 2/3 (B) -1.5 (C) 15% (D) 0.7
Solution:-
The probability of any event (E) always lies between 0 and 1 i.e. 0 ≤ P(E) ≤ 1. So, from the above options, option (B) -1.5 cannot be the probability of an event.
Q5. If P(E) = 0.05, what is the probability of ‘not E’?
Solution:-
We know that,
P(E) + P(not E) = 1
It is given that, P(E) = 0.05
So, P(not E) = 1 – P(E)
Or, P(not E) = 1 – 0.05
∴ P(not E) = 0.95
Q6. A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavored candy?
(ii) a lemon flavored candy?
Solution:-
(i) We know that the bag only contains lemon-flavored candies.
So, The no. of orange flavored candies = 0
∴ The probability of taking out orange flavored candies = 0/1 = 0
(ii) As there are only lemon flavored candies, P(lemon flavored candies) = 1 (or 100%)
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?
Solution:-
Let the event wherein 2 students having the same birthday be E
Given, P(E) = 0.992
We know,
P(E) + P(not E) = 1
Or, P(not E) = 1 – 0.992 = 0.008
∴ The probability that the 2 students have the same birthday is 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?
Solution:-
The total number of balls = No. of red balls + No. of black balls
So, the total no. of balls = 5 + 3 = 8
We know that the probability of an event is the ratio between the no. of favorable outcomes and the total number of outcomes.
=> P(E) = (Number of favorable outcomes/ Total number of outcomes)
(i) Probability of drawing red balls = P (red balls) = (no. of red balls/total no. of balls) = 3/8
(ii) Probability of drawing black balls = P (black balls) = (no. of black balls/total no. of balls) = 5/8
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?
(ii) white?
(iii) not green?
Solution:-
The Total no. of marbles= 5 + 8 + 4 = 17
P(E) = (Number of favorable outcomes/ Total number of outcomes)
(i) Total number of red marbles= 5
P (red marbles) = 5/17 = 0.29
(ii) Total number of white marbles= 8
P (white marbles) = 8/17 = 0.47
(iii) Total number of green marbles = 4
P (green marbles) = 4/17 = 0.23
∴ P (not green) = 1 – P (green marbles) = 1 – (4/17) = 0.77
Q10. A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹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 50 p coin?
(ii) will not be a ₹5 coin?
Solution:-
Total no. of coins = 100 + 50 + 20 + 10 = 180
P(E) = (Number of favorable outcomes/ Total number of outcomes)
(i) Total number of 50 p coin = 100
P (50 p coin) = 100/180 = 5/9 = 0.55
(ii) Total number of ₹5 coin = 10
P (₹5 coin) = 10/180 = 1/18 = 0.055
∴ P (not ₹5 coin) = 1 – P (₹5 coin) = 1 – 0.055 = 0.945
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. 15.4). What is the probability that the fish taken out is a male fish?
Solution:-
The total number of fish in the tank = 5 + 8 = 13
Total number of male fish = 5
P(E) = (Number of favorable outcomes/ Total number of outcomes)
P (male fish) = 5/13 = 0.38
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 Fig. 15.5), 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?
Solution:-
Total number of possible outcomes = 8
P(E) = (Number of favorable outcomes/ Total number of outcomes)
(i) Total number of favorable events (i.e. 8) = 1
∴ P (pointing at 8) = ⅛ = 0.125
(ii) Total number of odd numbers = 4 (1, 3, 5 and 7)
P (pointing at an odd number) = 4/8 = ½ = 0.5
(iii) Total numbers greater than 2 = 6 (3, 4, 5, 6, 7 and 8)
P (pointing at a number greater than 4) = 6/8 = ¾ = 0.75
(iv) Total numbers less than 9 = 8 (1, 2, 3, 4, 5, 6, 7, and 8)
P (pointing at a number less than 9) = 8/8 = 1
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