MQE 2013 QUESTION 23

[Ankit]

An urn contains 5 red balls, 4 black balls and 2 white balls. A player
draws 2 balls one after another with replacement. Then the probability
of getting at least one red ball or at least one white ball is:

(A) 105/121,
(B) 67/121,
(C) 20/121,
(D) None of the above

[ABC]

b.67/121

[Amit Goyal]

Ques: An urn contains 5 red balls, 4 black balls and 2 white balls. A player
draws 2 balls one after another with replacement. Then the probability
of getting at least one red ball or at least one white ball is:

(A) 105/121,
(B) 67/121,
(C) 20/121,
(D) None of the above

Ans: This is how you do it:
 Pr(at least one red ball or at least one white ball)
= 1 - Pr (both black balls)
= 1 - (4/11)(4/11)
= 105/121

[Sinistral]

what am I missing while counting???:

P =  P(both red balls) +    P(1R + 1B)    +  P(Both White)    +    P(1B + 1W)      +    P(1R + 1W)
   =    (5/11)(5/11)    +  (5/11)(4/11)  +     (2/11)(2/11)     +   (4/11)(2/11)     +   (5/11)(2/11)
   =     (25+20+4+8+10)/121
   =       67/121

[Sinistral]

ooops.. i just realised order is also important when choosing 1R + 1B, 1B + 1W, 1R + 1W balls. adding them twice will give the required no. 105. :D
*************************


what am I missing while counting???:

P =  P(both red balls) +    P(1R + 1B)    +  P(Both White)    +    P(1B + 1W)      +    P(1R + 1W)
   =    (5/11)(5/11)    +  (5/11)(4/11)  +     (2/11)(2/11)     +   (4/11)(2/11)     +   (5/11)(2/11)
   =     (25+20+4+8+10)/121
   =       67/121