Uses of Probability in Real Life
Probability related to playing cards: Now, let us take an example related to
The cards in each suit are ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2.
Kings, queens and jacks are called face cards.
4. One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will:
(i) be an ace,
(ii) not be an ace.
- Well shuffled cards mean equally likely outcomes.
- There are
aces in a deck. Let E be the event ‘the card is an ace’. Thus, number of outcomes favourable to E = while number of possible outcomes = - Therefore, P(E) =
4 52 - Let F be the event ‘card drawn is not an ace’. The number of outcomes favourable to the event F =
– = - Therefore, P(F) =
= - We have found the answers
Remark : Note that F is nothing but
5. Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match?
- Let S and R denote the events that Sangeeta wins the match and Reshma wins the match, respectively.
- The probability of Sangeeta’s winning = P(S) =
(given) - The probability of Reshma’s winning = P(R) = 1 – P(S) =
- We have found the answers
6. Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year).
- Say, Savita’s birthday can be any day of the year.Now, Hamida’s birthday can also be any day of 365 days in the year.
- Here, all
outcomes are equally likely. - (i) If Hamida’s birthday is different from Savita’s, the number of favourable outcomes is =
- So, P (Hamida’s birthday is different from Savita’s birthday) =
- P(Savita and Hamida have the same birthday) =
- We have found the answers
7. There are 40 students in Class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy?
- The number of all possible outcomes is
- The number of outcomes favourable for a card with the name of a girl =
- P(name of a girl) = P(Girl) =
25 40 - The number of outcomes favourable for a card with the name of a boy =
- P(card with name of a boy) = P(Boy) =
15 40 - We have found the answers
Note : We can also determine P(Boy), by taking P(Boy) = 1 – P(not Boy) = 1 – P(Girl) = 1-
8. A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be: (i) white? (ii) blue? (iii) red?
- Marble drawn at random means all the marbles are equally likely to be drawn.
- Number of possible outcomes =
- Let W denote the event ‘the marble is white’, B denote the event ‘the marble is blue’ and R denote the event ‘marble is red’.
- The number of outcomes favourable to the event W =
So, P(W) = - Similarly, P(B) =
= - P(R) =
- We have found the answers
Note: that P(W) + P(B) + P(R) = 1.
9. Harpreet tosses two different coins simultaneously (say, one is of Rs. 1 and other of Rs. 2). What is the probability that she gets at least one head?
- We write H for ‘head’ and T for ‘tail’. When two coins are tossed simultaneously,
- Possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all equally likely.
- Thus, the number of outcomes favourable to E is
. Therefore, P(E) = - The outcomes favourable to the event E, ‘at least one head’ are (H, H), (H, T) and (T, H).
- The probability that Harpreet gets at least one head is
3 4
Note : You can also find P(E) as follows: P (E) = 1 – P(
Since P(
Did you observe that in all the examples discussed so far, the number of possible outcomes in each experiment was finite?
There are many experiments in which the outcome is any number between two given numbers, or in which the outcome is every point within a circle or rectangle, etc.
Can you now count the number of all possible outcomes? As you know, this is not possible since there are infinitely many numbers between two given numbers, or there are infinitely
So, the definition of (theoretical) probability which you have learnt so far cannot be applied in the present form.
What is the way out? To answer this, let us consider the following example :
10. In a musical chair game, the person playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing. What is the probability that the music will stop within the first half-minute after starting?
- Let E be the event that ‘the music is stopped within the first half-minute’.
- Since all the outcomes are equally likely, the time period favourable to the event E is
- So, P(E) =
Time period favourable to the event E Time period in which outcomes can lie - P(E) =
1 2 2 - We have found the answer.
Now, let's solve another example involving probability of finding a crashed site.
11. A missing helicopter is reported to have crashed somewhere in the rectangular region with length 9 km and breadth 4.5 km. On the north-eastern corner, a rectangular lake with length 3 km and breadth 2.5 km is present. What is the probability that it crashed inside the lake?
- We know that the area of the rectangular region lake is:
where l - length and b - breadth of the considered region. - Using the formula we get: Area of the entire region where the helicopter can crash =
km 2 - Area of the lake =
km 2 - So, P (helicopter crashed in the lake) =
- We have found the answer.
12. A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that:
(i) it is acceptable to Jimmy?
(ii) it is acceptable to Sujatha?
- One shirt is drawn at random from the carton of 100 shirts. Therefore, there are
equally likely outcomes. - The number of outcomes favourable to Jimmy =
. Therefore, P (shirt is acceptable to Jimmy) = = - The number of outcomes favourable to Sujatha = (Number of good shirts)
+ (Number of shirts with minor defects) = - So, P (shirt is acceptable to Sujatha) =
= - We have found the answer.
Before proceeding to the next example, let's see what happens when two dice are rolled at a time.
Two dice are thrown at the same time. What is the probability that the sum of the two numbers appearing on top of the dice is 8.
1. E = sum of two numbers on top of dice is 8
This table shows all possible outcomes with X marking the favourable outcomes:
| X | 1 | 2 | 3 | 4 | 5 | 6 |
1 | ||||||
2 | X | |||||
3 | X | |||||
4 | X | |||||
5 | X | |||||
6 | X |
(2) Number of outcomes favourable to E =
(3) Number of all possible outcomes of the experiment=
So P(E) =
13. Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 8? (ii) 13? (iii) less than or equal to 12?
Note: The pair (1, 4) is different from (4, 1).
- The number of possible outcomes = First die outcomes × Second die outcomes = 6 × 6 =
. - Let event 'E' is favourable to the event ‘the sum of the two numbers is 8’. Thus, P(E) =
- The outcomes favourable to the event ‘the sum of the two numbers is 8’ denoted by E, are: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) i.e. the number of outcomes favourable to E = 5.
- Similarly, P(F) =
0 36 where the event F, ‘the sum of two numbers is 13’. - There is no outcome favourable to the event F.
- Let event G denoted outcomes favourable to ‘sum of two numbers is 12’. Thus, P(G) =
= - All the outcomes are favourable to the event G, ‘sum of two numbers is 12’.
- We have found the answer.