The probability that a company executive will travel by plane is (2/5)

1.	The probability that a company executive will travel by plane is (2/5) and that he will travel by train is (1/3). Find the probability of his travelling by plane or train.
Answer» let A denote the event that a company executive will travel by plane and B denote the event of him travelling by train Given : P(A) = \(\frac{2}{5}\), P(B) = \(\frac{1}{3}\) To find : Probability of a company executive will be travelling by plane or train = P(A or B) Formula used : P(A or B) = P(A) + P(B) - P(A and B) Probability of a company executive will be travelling in both plane and train =P(A and B)= 0 (as he cannot be travelling by plane and train at the same time) P(A or B) = \(\frac{2}{5}\) + \(\frac{1}{3}\) – 0 P(A or B) = \(\frac{6+5}{15}\) = \(\frac{11}{15}\) P(A or B) = \(\frac{11}{15}\) Probability of a company executive will be travelling by plane or train= P(A or B) = \(\frac{11}{15}\)

The probability that a company executive will travel by plane is (2/5) and that he will travel by train is (1/3). Find the probability of his travelling by plane or train.

Answer»

let A denote the event that a company executive will travel by plane and B denote the event of him travelling by train

Given : P(A) = \(\frac{2}{5}\), P(B) = \(\frac{1}{3}\)

To find : Probability of a company executive will be travelling by plane or train = P(A or B)

Formula used : P(A or B) = P(A) + P(B) - P(A and B)

Probability of a company executive will be travelling in both plane and train =P(A and B)= 0

(as he cannot be travelling by plane and train at the same time)

P(A or B) = \(\frac{2}{5}\) + \(\frac{1}{3}\) – 0

P(A or B) = \(\frac{6+5}{15}\) = \(\frac{11}{15}\)

P(A or B) = \(\frac{11}{15}\)

Probability of a company executive will be travelling by plane or train= P(A or B) = \(\frac{11}{15}\)

Discussion

No Comment Found

Related InterviewSolutions

If A and B are independent events., where P(A) = 0.3, P(B) = 0.6, then find: (i) P(A ∩ B) (ii) P(A ∪ \(\bar{B}\)) (iii) P(A ∪ B) (iv) P(\(\bar{A}\) ∩ \(\bar{B}\))
Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?
A combination lock on a suitcase has 3 wheels, each labelled with nine digits from 1 to 9. If an opening combination is a particular sequence of three digits with no repeats, what is the probability of a person guessing the right combination ?
The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?
A couple has 2 children. find the probability that the both are boys if it is given that (i) one of the child is boy (ii) the older child is boy.
The probability of selecting a green marble at random from a jar that contains green,white and yellow marble is 1/3. The probability of selecting a white marble random from the jar is 2/9.If the jar contains 8 yellow marbles, find the total numbers of marbles in the jar
A bag contains 3 red, 3 white and 3 green balls. One ball is taken out of the bag at random. What is the probability that the ball drawn is:i. red. ii. not red. iii. either red or white.
In a hockey team there are 6 defenders, 4 offenders and 1 goalee. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection thatA. (1) the goalee will be selectedB. a defender will be selected.C.D.
In a hockey team there are 6 defenders , 4 offenders and 1 goalie. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection that: i. The goalie will be selected. ii. A defender will be selected.
If n(A) = 2, P(A) = 1/5, then n(S) = ? (A) 10 (B)5/2(C) 2/5(D) 1/3