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1.A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is i) 3 ii) 12.
The coin with 1 marked on one face and 6 on the other face.
The coin and die are tossed together.
. . . S = { ( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) , ( 1 , 4 ) , ( 1 , 5 ) , ( 1 , 6 ) , ( 6 , 1 ) , ( 6 , 2 ) , ( 6 , 3 ) , ( 6 , 4 ) , ( 5 , 5 ) , ( 6 , 6 ) }.
i)Let A be the event having sum of numbers is 3
Thus P (A) = 1 /12 .
ii)Let B be the event having sum of number is 12.
. . . P (B) = 1 / 12 .
2.There are four men and six women on the city council. If one council member is selected for a committee at random , how likely is it that it is a woman?
Here total members in the council 1 = 4 + 6 = 10
one member is selected out of 10 members.
n (S) = 10 C1 = 10
Let A be the event that the member is a woman
. . . n (A) = 6 C1 = 6 .
3.If 2 / 11 is the probability of an event , what is the probability of the event ' not A' .
Here probability of event A is P ( A ) = 2 / 11
. . . probability of event not A is P (A¯) = 1- P(A) = 1 - 2 / 11 = 9 / 11 .
4.A letter is chosen at random from the word ASSASSINATION : Find the probability that letter is
i) a vowel ii) a consonant
There are 13 letters in the word ASSASSINATION which contains 6 vowels and 7 consonants.
One letter is selected out of 1 letters in 13C1 = 13 ways.
i)One vowel is selected out of 6 vowels in 6C1 = 6 ways
Thus probability of a vowel = 6 / 13 .
ii)One consonant is selected out of 7 consonants in 7C1 = 7 ways.
Thus probability of consonant = 7 / 13 .
5.In a lottery , a person chosen six different natural numbers at random from 1 to 20 and if these size numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?
Number of numbers in the draw = 20
Number of numbers to be selected = 6
Let Δ be the event that six numbers match with the six numbers fixed by the lottery committee.
. . . n (A) = 6C6 = 1
Thus probability of winning the prize P (A)
6.Check whether the following probabilities P (A) ans P (B) are consistently defined
i)P (A) = 0.5 , P (B) = 0.7 , P (A∩B) = 0.6
ii)P (A) = 0.5 , P (B) = 0.4 , P (AυB) = 0.8
i) Here P (A) = 0.5 , P (B) = 0.7 and P (A∩B) = 0.6
Now P (A∩B) > P (A)
Thus the given probabilities are not consistently defined.
ii)Here P (A) = 0.5 , P (B) = 0.4 and P (AυB) = 0.8
We know that P (AυB) = P (A) + P (B) - P (A∩B)
0.8 = 0.5 + 0.4 - P (A∩B)
P(A∩B) = 0.9 - 0.8 = 0.1
. . . P ( A∩B) < P (A) and P (A∩B) < P (B)
Thus the given probabilities are consistently defined.
7.Given P (A) = 3 / 5 and P (B) = 1 / 5 . Find P (A or B ), if A and B are mutually exclusive events.
Here P (A) = 3 / 5 , P (B) = 1 / 5.
Since A and B are mutually exclusive events
P (AυB) = P (A) + P (B)
. . . P (AυB) = 3 / 5 + 1 / 5 = 4 / 5 .
8.If E and F are events such that P (E) = 1 / 4 , P (F) = 1 / 2 and P (E and F ) = 1 / 8 . Find
i) P ( E and F ) ii) P ( not E and not F )
Here P ( E ) = 1 / 4 , P ( F ) , P ( F ) = 1 / 2 and P ( E ∩F ) = 1 / 8 .
i) We know that P ( EυF ) = P ( E ) + P ( F ) - P ( E∩F )
9.Events E and F are such that P (not E or not F) = 0.25. state whether E and F are mutually exclusive.
Here P (not E or not F) = 0.25
Thus E and F are not mutually exclusive events.
10.A and B are events such that P (a) = 0.42 , P (b) = 0.48 and P ( A and B ) = 0.16. Determine
i) P (not A) ii) P ( not B ) and iii) P (A or B).
11.In class XI of a school , 40% of the students study Mathematics and 30% study Biology, 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.
Let A be the event that the student is studying Mathematics and B be the event that the student is studying biology.
12.In an entrance test that is graded on the basis of two examination , the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7 . The probability of passing at least one of them is 0.95. What is the probability of passing both?
Let A be the event that the student passes the first examination and B be the event that the student passes the second examination.
Then P (A) = 0.8 , P (B) = 0.7 and P (AυB ) = 0.95
We know that P (AυB) = P (A) + P (B) - P (A∩B)
0.95 = 0.8 + 0.7 - P (A∩B)
P (A∩M) = 1.5 - 0.95 = 0.55.