Consider a scenario that you when to a shop where you are the fourth person in the line to purchase the item (you only want to purchase bottle), shop keeper only has two bottle and one ice cream left. All four person do not have any clue that shop keeper only left with few items. Thus, there is an equal chance that everyone can either for bottle or ice cream. Everyone in the line planned to purchase only one item.
Now answer the following questions.
Solution:
Given
The number of bottles left with shop keeper =2
The number of ice cream left with shop keeper =1
The position in the line = 4
The probability that a person buy ice cream =1/2
The probability that a person buy bottle =1/2
The possible ways of getting bottle to fourth person is only when the first three person wish to buy these
a. All three wish to buy ice cream
b. Two wish to buy ice cream and one bottle
The probability that all three wish to buy ice cream =1/2×1/2×1/2=1/8
The probability that two wish for ice cream and one wish for bottle=1/2×1/2×1/2=1/8
The possible combinations of two wish for ice cream and one wish for bottle are shown below
a) ice cream, ice cream, bottle
b)ice cream, bottle, ice cream
c)bottle, ice cream, ice cream
Thus they can be arranged in three ways.
The total probability that two wish for ice cream and one wish for bottle =3×1/2×1/2×1/2=3/8
(1) The probability that you will get bottle in fourth position in line=1/8 + 3/8 = 1/2
(2) Here is the tree diagram:
Here if we observe that each person has two choices to make and probability of each choice remain same for all people and there are four person in line and choice of each person is independent of others. This is very similar to condition that there are two possibilities and their probability remain same and there are fixed number of trails each trail is independent. This is binomial probability distribution.
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