(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 __________.
(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
(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 a solution: a true-false question. The solution is right or wrong.
(iv) A baby is born. It is a boy or a girl.
(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) 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.
The tossing of a coin is a fair way of deciding because the number of possible outcomes is only 2, i.e. either head or tail. Since these two outcomes are equally likely outcomes, tossing is unpredictable and is considered to be completely unbiased.
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.
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
(i) an orange-flavoured candy?
(ii) a lemon-flavoured candy?
(i) an orange-flavoured candy?
We know that the bag only contains lemon-flavoured candies.
So, The number of orange-flavoured candies = 0
The probability of taking out orange-flavoured candies = $\dfrac{0}{1}$ = 0
(ii) a lemon-flavoured candy?
As there are only lemon-flavoured candies, P(lemon-flavoured candies) = 1 (or 100%)
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
(i) red?
(ii) not red?
The total number of balls = No. of red balls + No. of black balls
So, the total number of balls = 5+3 = 8
We know that the probability of an event is the ratio between the number of favourable outcomes and the total number of outcomes.
| P(E) = $\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes}$ |
(i) red?
Probability of drawing red balls = P (red balls) = $\dfrac{no.\: of \: red \: balls}{total \: no. \: of \: balls}$ = $\dfrac{3}{8}$
(ii) not red?
(ii) Probability of drawing black balls = P (black balls) = $\dfrac{no. \: of \: black \: balls}{total \: no. \: of \: balls}$ = $\dfrac{5}{8}$
(i) red?
(ii) white?
(iii) not green?
The Total number of balls = 5+8+4 = 17
| P(E) = $\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \:outcomes}$ |
(i) red?
Total number of red balls = 5
P (red ball) = $\dfrac{5}{17} = 0.29$
(ii) white?
Total number of white balls = 8
P (white ball) = $\dfrac{8}{17} = 0.47$
(iii) not green?
Total number of green balls = 4
P (green ball) = $\dfrac{4}{17} = 0.23$
P (not green) = 1-P(green ball) = 1-$\dfrac{4}{7} = 0.77$
(i) will be a 50 p coin?
(ii) will not be a ₹5 coin?
Total no. of coins = 100+50+20+10 = 180
| P(E) = $\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes}$ |
(i) will be a 50 p coin?
Total number of 50 p coin = 100
P (50 p coin) = $\dfrac{100}{180} = \dfrac{5}{9} = 0.55$
(ii) will not be a ₹5 coin?
Total number of ₹5 coin = 10
P (₹5 coin) = $\dfrac{10}{180} = \dfrac{1}{18} = 0.055$
P (not ₹5 coin) = 1-P (₹5 coin) = 1-0.055 = 0.945