Probability – Independent Events

INDEPENDENT EVENTS

Events are independent if the occurrence of either of them does not affect the probability that the others occur.

Consider again the example on the previous page. Suppose we happen to choose a blue ball from box X. This does not affect the outcome when we choose a ball from box Y. The probability of A selecting a red ball from box Y is ¾ regardless of which colour ball is selected from box X.
So, the two events “a blue ball from X” and “a red ball from Y” are independent.

If A and B are independent events then P(A and B)=P(A)×P(B). This rule can be extended for any number of independent events.

For example:
If A, B, and C are all independent events, then P(A and B and C) =P(A)×P(B)×P(C).

Q1. A coin and a die are tossed simultaneously. Determine the probability of getting a head and a 3 without using a grid.
Solution:

P(a head and a 3) =P({H})×P(3) {events are independent}
=½×⅙=1/12

Q2. There are two identic bags, first one with 4 red and 6 white balls and the second with 1 red and 5 white balls. If one ball is picked from either of the two bags, then what is the probability that the ball picked is red?
a. 0.28 b. 0.31 c. 0.4 d. 0.2
correct: a, solution:
Do two these steps in a row, choose one bag then pick a ball inside. Probability of selecting any bag is 0.5. The required probability

=½×4/10+½×⅙=17/60=0.28

Q3. Two identical bags are filled with balls. Bag A contains 3 pink and 2 yellow balls. Bag B contains 5 pink and 4 yellow balls. It is equally likely that Bag A or Bag B is chosen. Each ball has an equal chance of being chosen from the bag. A bag is chosen at random and a ball is then chosen at random from the bag.
a) Represent the information by means of a tree diagram. Clearly indicate the probability associated with each branch of the tree diagram and write down all the outcomes.
b) What is the probability that a yellow ball will be chosen from Bag A?
c) What is the probability that a pink ball will be chosen?
solution:
a)

Q4. Probability that Chaman will move one step forward is 0.4 and probability that he will move one step backward is 0.6 What is the probability that Chaman will be at his original position after moving two steps?
a. 0.48 b. 0.52 c. 0.16 d. 0.36
correct: a, solution:
The chance that he’ll be at his original place =P(that he moves forward and then backwards)+P(that he moves backwards and then forward)
=2×0.4×0.6=0.48

Or:
Chaman will move either (one step forward first then one step backward) or (one step backward first then one step forward).
(0.4×0.6)+(0.6×0.4)

Q5. In throw of a die and toss of a coin, what is the probability of getting either 6 on the die or tail on the coin or both?
a. 5/12 b. ⅔ c. 7/12 d. ¾
correct: c, solution:

P(A)=P(6 on the dice) =⅙.
P(B)=P(Tails on the coin) =½.
P(A∩B)=P(both)
P(A∪B)=P(A)+P(B)-P(A∩B)

Because there are two independent events then
P(A∩B)=P(A)×P(B)
=⅙+½-(⅙×½)=7/12

Q6. It is given that two events, A and B, are independent. P(A)=⅖ and P(B)=0.35. Calculate P(A or B).
Solution:

P(A and B)=P(A)×P(B)=0.4×0.35=0.14
P(A or B)=P(A)+P(B)-P(A and B)
=0.4+0.35-0.14=0.61

Q7. Determine whether the two events are independent or dependent.
1. Tossing a coin and selecting a card from a deck
2. Driving on ice and having an accident
3. Drawing a ball from an urn, not replacing it, and then drawing a secondball
4. Having a high I.Q. and having a large hat size
5. Tossing one coin and then tossing a second coin
solution:
1. Independent.Tossing a coin has no effect on drawing a card.
2. Dependent. In most cases, driving on ice will increase the probabilityof having an accident.
3. Dependent. Since the first ball is not replaced before the second ballis selected, it will change the probability of a specific second ballbeing selected.
4. Independent. To the best of the author’s knowledge, no studies havebeen done showing any relationship between hat size and I.Q.
5. Independent.The outcome of the first does not influence the outcomeof the second coin.

Q8. Probability of solving specific problem independently by A and B are ½ and ⅓ respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved (ii) exactly one of them solves the problem.
Probability of solving the problem by A, P(A)=½
Probability of solving the problem by B, P(B)=⅓
Since the problem is solved independently by A and B,

∴P(A∩B)=P(A)∙P(B)=½×⅓=⅙
P(Ā)=1-P(A)=1-½=½
P(B̄)=1-P(B)=1-⅓=⅔

(i) Probability that the problem is solved=P(A∪B)
P(A∪B)=P(A)+P(B)-P(A∩B)
P(A∪B)=½+⅓-⅙=⅔

(ii) Probability that exactly one of them solves the problem is given by,
P(A)∙P(B̄)+P(Ā)∙P(B)
½×⅔+½×⅓=½(1)=½