More applications of Probability
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.
If we roll a die, the result is a number between 1 and 6, and all outcomes are equally likely. If we roll two dice at once and add up their scores, we can get results from
Some results can only happen one way (to get 12 you have to roll + ) while others can happen in multiple different ways (to get 5 you have to roll either + or + ).
This table shows all possible outcomes:
2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
The most likely result when rolling two dice is 7. There are
The least likely outcomes are 2 and 12, each with a probability of
It is impossible to forecast the outcome of a single coin toss or die roll. However, using probability we can very accurately predict the outcome of
If we throw a die 30 times, we know that we would get around
If we roll it 300 times, there will be around
In this animation you can roll many “virtual” dice at once and see how the results compare to the predicted probabilities:
Rolling Dice
We roll
Notice how, as we roll more and more dice, the observed frequencies become closer and closer to the frequencies we predicted using probability theory. This principle applies to all probability experiments and is called the law of large numbers.
Similarly, as we increase the number of dice rolled at once, you can also see that the probabilities change from a straight line (one die) to a triangle (two dice) and then to a “bell-shaped” curve. This is known as the central limit theorem, and the bell-shaped curve is called the normal distribution.