Let A = Event that A speaks the truth
B = Event that B speaks the truth
Then P(A) = 75/100 = 3/4
P(B) = 80/100 = 4/5
P(A-lie) = = 1/4
P(B-lie) = = 1/5
Now, A and B contradict each other =[A lies and B true] or [B true and B lies]
= P(A).P(B-lie) + P(A-lie).P(B)
=
= = 35%
Total number of elementary events =
Given,third ticket =30
=> first and second should come from tickets numbered 1 to 29 = ways and remaining two in ways.
Therfore,favourable number of events =
Hence,required probability = =551 / 15134
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
P(E) =n(E)/n(S)=7/8.
Possible outcomes = (RY, YP, PR)
2C1 4C1 + 4C1 6C1 + 6C1 2C1
Required probability = (2C1 4C1 + 4C1 6C1 + 6C1 2C1)/(12C2)
= 8 + 24 + 12/66
= 44/66
= 2/3.
P(first letter is not vowel) =
P(second letter is not vowel) =
So, probability that none of letters would be vowels is =
A red ball can be drawn in two mutually exclusive ways
(i) Selecting bag I and then drawing a red ball from it.
(ii) Selecting bag II and then drawing a red ball from it.
Let E1, E2 and A denote the events defined as follows:
E1 = selecting bag I,
E2 = selecting bag II
A = drawing a red ball
Since one of the two bags is selected randomly, therefore
P(E1) = 1/2 and P(E2) = 1/2
Now, = Probability of drawing a red ball when the first bag has been selected = 4/7
= Probability of drawing a red ball when the second bag has been selected = 2/6
Using the law of total probability, we have
P(red ball) = P(A) =
=
We know that,
Total number of balls n(S) = 26
Number of vowels n(E) = 5
Hence, required probability = n(E)/n(S) = 5/26.
3 letters can be choosen out of 9 letters in ways.
More than one vowels ( 2 vowels + 1 consonant or 3 vowels ) can be choosen in ways
Hence,required probability = = 17/42
One person can select one house out of 3= ways =3.
Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9.
Therefore, probability that all thre apply for the same house is 1/9
P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7
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