Probability Quiz – 1

Let P(L) = probability of selecting a leap year. P(NL) = probability of selecting a non - leap year. P(M) = probability of getting 53 Mondays in a year.
P(M)=P(M|L)P(L)+P(M|NL)P(NL) (By law of probability)
=\frac{2}{7}\times \frac{1}{4}+ \frac{1}{7}\times \frac{3}{4}=\frac{5}{28}

Let A be the event that the selected bottle is flat, X be the event that it is chosen from carton 1, Y be the event that it is chosen from carton 2, Z be the event that it is chosen from carton 3.
P(A)=P(A|X)P(X)+P(A|Y)P(Y)+P(A|Z)P(Z)
=\frac{4}{10}\times \frac{1}{3}+\frac{1}{6}\times \frac{1}{3}+\frac{3}{8}\times \frac{1}{3}=\frac{113}{360}

Since, A{4,5,6} and B={1,2,3,4}
A⋂B={4}
P(A∪B)=P(A)+P(B)-P(A⋂B)
P(A∪B)=\frac{3}{6}+\frac{4}{6}-\frac{1}{6}=1

Let A be the event that sum of the digits on the selected ticket is 8, then
Let B be the event that the product of the digits is zero.
B={00, 01, 02, 03 ….., 09, 10, 20, 30, 40}
(A⋂B) = {8}
Required probability = P(\frac{A}{B})
=P\frac{(A⋂B)}{(B)}
=\frac{1}{14}

We know that the sum of two sides in a triangle is greater than the third side. Therefore, in following 3 cases, the triangle is not formed.
2. 3, 5
2, 3, 6
2, 4, 6
Total number of triangles formed using 5 lines = ^{5}C_{3}
Therefore, required probability \frac{3}{^{5}C_{3}}=\frac{3}{10}

he probability of a boy being selected = \frac{7}{15}. Therefore, the probability of having either a pencil or an eraser with him = \frac{7}{15}\times\frac{10}{15}=\frac{14}{45}

Chhota Chetan knows that the code is a three digit number. Total number of numbers that can be formed = 3x(9x9x1) = 243. Of these only one number will open the locker.
The probability of his success = 1/243.

5 can be obtained by getting any of [(4, 1), (3, 2). (2. 3). (1. 4)]
7 can be obtained by getting any of [(6, 1). (5, 2). (4. 3). (3. 4). (2. ‘5). (1. 6)]
Probability that 5 comes before 7 = 4/10 = 0.4

Probability of getting 7 before 5 = \frac{6}{4+6}=0.6

Out of numbers 00, 01, 02,…., 99, those numbers the product of whose digits is 18 are 29,36, 63, 92 i.e., only 4.
p=P(E)=\frac{4}{100}=\frac{1}{25},q=P(E)=1-\frac{1}{25}=\frac{24}{25}
Let X be the random variable, showing the number of times E occurs in 4 selections. Then p(so that E occurs at least 3 times) = p(X = 3 or X = 4)
=p(x=3)+p(x=4)= 4C_{3}p^{3}q^{1}+4C_{4}p^{4}q^{0}=4p^{3}q+p^{4}
= 4\times(\frac{1}{25})^{3}\times\frac{24}{25}+(\frac{1}{25})^{4} = \frac{97}{390625}