Probability - 4

Dumb question - Why probability of P(A/B₀) and P(A/B₁) is zero ?

Ans - Because in event we are getting 2 white ball.

and when in sample we have none or one white we can not get 2 white ball.

Stockist approach :-

Q- A bag contains 5 fair coin, 2 doulaly headed coin, 2 doubly tailed and are biased coin. Probability of head with biased = 2/3 . A coin is selected and tossed. If head appeared same coin is tossed and if tail appears another coin selected from remaining and tosed. Find the chasce -

(1) Same .coin is tessed twice .

(2) Chance .head appears in both toss.

(3)If head appears in both wins. find the chance that it is doubly headed coin .

Ans - Total coin = 5 + 2 + 2 + 1 = 10



(1) P(getting H in I ^st toss) =

(2) P(getting both (H H) in 2 toss) =

(3) P(Doubly head / H H )

=

In finite samble space (Geometrical probaility ) :-

(1) If a point is randomly sclected from area 'S' which indudes an area 6 then chance that it is sclected from area 6





(2) If a point is randomly taken from line AB of length L then chance that it is selected from the segment PQ(l) contained by AB is = l/2

Illustration :- In a circle of radius 'a'. Find the probalility that point is close to the circumference than to its centre.

Ans - n(S) =

E : Point is close to circumference than oits centre.

P(E) =



Dumb question - How probability of " point is close to circumference than its centre " is 3/4 .

Ans - A point is close to circumference than its centre when it lies in region of outside the circle of r = a/2 and inside the circle of r = a .

P(E) =



Easy level
Q. 1 - A natural no. is randomly selected. Find the chance that digit in unit place of its square is 4.

Ans :-

There are only two digit (2 and 8). Whose square's unit place is 4.
So, P(E) = 2/10 = 1/5

Q. 2 4 red and 3 white ball are arranged | n a row. Find chance that two ball at extreme are whit .

Ans :- Case I:- alike balls



n(S)=

n(A)=

P(A)=

Case2:- Different balls-

n(S) = 7!

n(A) = 3C₂. 2!, 5!

P(A) =

Q. 3- P(A) = 1/2 ,P(B) = 1/2 . Find the least and greatest valume of intersection P(A img B).

Ans - P(A B) = P(A). P(B/A)
P(A B) greatest = 1/2 if P(B/A) = 1

=



Q. 4- A and B are independent then show that and are also independent .

Ans - Note - If and are independent. then P(A) . P(B)
T.PT :-
L. H. S. :- { 1 - P(A) } { 1 - P(B)}

= 1 - P(A) - P(B) + P(A) . P(B)
= 1 - [P(A) + P(B) - P(A B)]
= 1 - P(A B)
= P ()

Q. 5 - Two cords are drawn from 52 cards pack. Find the chance that both the cards are aces.

Ans- n(S) = no. of ways drawing 2 cards from 52 cards.

= 52 C₂

n(A) = no. of ways drawing 2 ace cards from 4 ace card

= 4 C₂

P(E) =

P(E) =