# 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 = p

^{2}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)

+ P

^{2}= 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 C

_{4}

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__- E

_{1}I

^{st}lot is chosen

E

_{2}2

^{nd}lot is chosen

E lot is rejected

P(E) = P(E

^{1}) P(E/E

_{1}) + P(E

_{2}). P(E/E

_{2})

P(E

_{1}) = P(E

_{2}) = 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

^{st}draw, + 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 .

B

_{1}: . letter is from

B

_{2}: . letter is from

B

_{3}: . letter is from

P(B

_{1}) = P(B

_{2}) = P(B

_{3}) =

P(B/B

_{1}) = =

P(A/B

_{2}) = 1/6. P(A/B

_{2}) = 1/5.

P(A) = P(B

_{1}). P(A/B

_{1}) + P(B

_{2})+ P(A/B

_{2}) + P(B

_{3}) P(A/B

_{3})

=

P (B

_{1}/ A) =

P(B

_{1}/ A) =