Probability - 6

Medium level

Q. 1- A, B and C are three event such that -

P(exatly one A or B) = P(exactly one B C)

=> P(exactly one C or A) = P

and = p² and A, B and C are
exhaustive. Find = value of P .

Ans - Not :- In exhaustive events

= 1

P = P (a) + P (B) - 2P ( A B )

P = P (B) + P (C) - 2P ( B C )

P = P (C) + P (A) - 2P ( C A )

3 P = 2 (P(A) + P(B) + P(C)) - 2P(A B) + P(A B) + P(C A))

P(A) + P(B) + P(C)[ ( A B) + P(B C) +P(C A) ]

+ [ P(A B) + P(B C) +P(C A) ] P(A) + P(B) + P(C) ...........(1)


P(A B C) = P(A) + P(B) + P(C) - [ P(A B) + P(B C) + P(C A) + P(A B C)........(2)

+ P (ABC) = P(ABC)

If A, B and C are exhaustive .

then P(A BC)=P(A) + P(B) + P(C) holds -

From eqn ........... (2)

P(A B C) = P(A B) + P(B C) + P(C A)


P utting in eqn ............ (1)

+ P(ABC) = P(ABC)

+ P² = 1 ..............(3)

On solving

P =

Q. 2. 12 face cards are removed from 52 cards and remaining 40 cards are well shuffled and 4 cards are drawn from pack of 40 cards. Find the chance that all four are of different suit and different denomination .

Ans- Heart Diamond club spade

n(s) = 40 C₄

No. of way drawing I^st card = 40 ways

No. of way drawing 2^nd card = 27 ways

No. of way drawing 3^rd card = 16 ways

No. of way drawing 4^th card = 7 ways

Total no. of way of drawing all 4 cards = 40 X 27 X 16 X 7

Dumb question - Why NO. of ways of drawing 2^nd card is 27 ?

Ans- Let I^st card drawn is ace of Heart. So, no, Heart and no ace card is further selected. So remaining cards are (40 - 10 - 3) = 27
So, 27 ways

P(E)= i

Q. 3 - There are two luts of article one lot contains 3 defective and 5 good article and other lot contains 4 defective and 8 good articles. A lot is randomly seleste a and 3 articles are drawn . this lot will be rejected if 2 or mor than 2 article are found to be defective. Find the chance that this be rejected.

And - E₁ I^st lot is chosen

E₂ 2^nd lot is chosen

E lot is rejected

P(E) = P(E¹) P(E/E₁) + P(E₂). P(E/E₂)

P(E₁) = P(E₂) = 1/2 (equally likely events)

P(E) +



Q. 4 A boy contains | coin of worth M rupees and n coins of total . Worth of M rupees. Coins are drawn without replace -ment till coin whose valime is M, is drawn, I Find the expectaion of draw .

Ans - P(drawing M in I^st draw) =

P(drawing M in 2^nd) drawn) = other coin draw, drawing of M)
P(drawing M in (r + 1)^th draw) =

Expectation = M coin draw In I^stdraw, + Mcoin draw in II^nd draw , + In 3^rd draw


=

=

Expectation =

Q. 5 - A letter is known to khave either from Agra or from Mathura or. Satara. on the stamp two consecultive word 'RA'are legible. Find the chance that letter from Agra .

Ans- A : .Two consecutive word 'RA'are legible .

B₁: . letter is from

B₂: . letter is from

B₃: . letter is from

P(B₁) = P(B₂) = P(B₃) =

P(B/B₁) = =

P(A/B₂) = 1/6. P(A/B₂) = 1/5.

P(A) = P(B₁). P(A/B₁) + P(B₂)+ P(A/B₂) + P(B₃) P(A/B₃)

=

P (B₁ / A) =



P(B ₁ / A) =