Example 1 : A coin is tossed 1000 times with the following frequencies: Head : 455,Tail : 545.Compute the probability for each event.
Solution : Since the coin is tossed 1000 times, the total number of trials is 1000. Let’s Call the events of getting a head and of getting a tail as E and F, respectively. Then, the number of times E happens, i.e., the number of times a head comes up, is 455.
So,the probability of E = Number of heads/Total number of trials
i.e.,P (E) = 455/1000 = 0.455
Similarly, the probability of the event of getting a tail F = Number of tails/Total number of trials
i.e.,P(F) =545/1000 = 0.545
Note that in the example above, P(E) + P(F) = 0.455 + 0.545 = 1, and E and F are the only two possible outcomes of each trial.
Example 2 : A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3,4, 5 and 6 as given in the following table
Outcome | 1 | 2 | 3 | 4 | 5 | 6 |
Frequency | 179 | 150 | 157 | 149 | 175 | 190 |
Find the probability of getting each outcome.
Solution : Let Ei denote the event of getting the outcome i, where i = 1, 2, 3, 4, 5, 6.Then Probability of the outcome 1 =P(E1)=Frequency of 1/Total number of times the die is thrown
= 179/1000=0.179
Similarly,P(E2) =150/1000 = 0.15,
P(E3) = 157/1000 = 0.157,
P(E4) =149/1000 = 0.149,
P(E5) = 175/1000 = 0.175
And P(E6) =190/1000 = 0.19
Note that P(E1) + P(E2) + P(E3) + P(E4) + P(E5) + P(E6) = 1
Also note that:
(i)The probability of each event lies between 0 and 1.
(ii)The sum of all the probabilities is 1.
(iii)E1, E2, . . ., E6 covers all the possible outcomes of a trial.
Example 3 : There are 6 pillows in a bed, 3 are red, 2 are yellow and 1 is blue. What is the probability of picking a yellow pillow?
Solution : The probability is equal to the number of yellow pillows in the bed divided by the total number of pillows, i.e. 2/6 = 1/3.
Example 4 : Fifty seeds were selected at random from each of 5 bags of seeds, and were kept under standardised conditions favourable to germination. After 20 days, the number of seeds which had germinated in each collection were counted and recorded as follows-
Bag | 1 | 2 | 3 | 4 | 5 |
No. of seeds germinated | 40 | 48 | 42 | 39 | 41 |
What is the probability of germination of
(i)more than 40 seeds in a bag?
(ii)49 seeds in a bag?
(iii)more than 35 seeds in a bag?
Solution : Total number of bags is 5.
(i) Number of bags in which more than 40 seeds germinated out of 50 seeds is 3.
P(germination of more than 40 seeds in a bag) = 3/5 = 0.6
ii)Number of bags in which 49 seeds germinated = 0.
P(germination of 49 seeds in a bag) = 0/5 = 0
(iii)Number of bags in which more than 35 seeds germinated = 5.So, the required probability = 5/5 = 1
In all the above examples you can observe that Probability can range in between 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.