Aricent Probability Quiz-1

|X| can take 7 values.
To get |X|<2 ( i.e., −2 ⇒ P(|X|<2)= Favourable CasesTotal Cases
= 3/7

Given, A be the event that X is selected and B is the event that Y is selected.

P(A)=17, P(B)=29.

Let C be the event that both are selected.

P(C)=P(A)×P(B) as A and B are independent events:

=(17)×(29)
= 2/63

P(Both are red),

=6C213C2
=5/26

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:

=6×6×6×6=64
n(S)=64
Let X be the event that all dice show the same face.

X={(1,1,1,1,),(2,2,2,2),(3,3,3,3),(4,4,4,4),(5,5,5,5),(6,6,6,6)}
n(X)=6
Hence required probability,

=n(X)n(S)=664
=1/216

The probability that first ball is white:

=12C130C1
=1230
=25
Since, the ball is not replaced; hence the number of balls left in bag is 29.

Hence, the probability the second ball is black:

=18C129C1
=1829
Required probability,

=(25)×(1829)
=36145

The probability that first toy shows the even number,

=1021
Since, the toy is not replaced there are now 9 even numbered toys and total 20 toys left.

Hence, probability that second toy shows the even number,

=920
Required probability,

=(1021)×(920)
=942

A number divisible by 4 formed using the digits 1, 2, 3, 4 and 5 has to have the last two digits 12 or 24 or 32 or 52.

In each of these cases, the five digits number can be formed using the remaining 3 digits in 3!=6 ways.

A number divisible by 4 can be formed in 6×4=24 ways.

Total number that can be formed using the digits 1, 2, 3, 4 and 5 without repetition

=5!=120
Required probability,

=24/120
=1/5

Let A be the event that ball selected from the first bag is red and ball selected from second bag is green.

Let B be the event that ball selected from the first bag is green and ball selected from second bag is red.

P(A)=(5/8)×(6/10)
=3/8 and

P(B)=(3/8)×(4/10)
=3/20
Hence, required probability,

=P(A)+P(B)
=3/8+3/20
=21/40

Different possible cases of contradiction,

A speaks truth and B does not speaks truth.

Or, A does not speak truth and B speaks truth.

=(3/4×1/5)+(1/4×4/5)
=3/20+4/20
=7/20
=35%

Let X be the event that cards are in a club which is not king and other is the king of club.

Let Y be the event that one is any club card and other is a non-club king.

Hence, required probability:

=P(A)+P(B)
=12C1×1C152C2+13C1×3C152C2
=(2×(12×1)52×51)+(2(13×3)52×51)
=(24+7852×51)
=1/26