# Solved Example #5

February 9th, 2013 by

Question : Find the probability that a rectangle selected at random from a 10*10 square is a square? a.17/55 b.7/55 c.9/55 d.16/55   Answer :  Total number of rectangles = 11C2 * 11C2 = 55*55 Total number of squares = 1^2 + 2^2 ....... 10^2 = 10*11*21/6 = 5*11*7 = 55*7 Required probability = 55*7 / 55*55 = 7/55   Crack CAT with Unacademy! Use referral code HANDA to get 10% off. Daily Live Classes Live Tests and Quizzes Structured Courses Personalized Coaching Enroll Now Download Unacademy App

# Solved Example #4

February 9th, 2013 by

Question : There are 4 red & 5 green balls in bag A and 5 red & 6 green balls in bag B. If a bag is selected at random and a ball is selected from that, what is the probability that it is red?   Answer :  Probability of Red Ball = Bag A is selected & Red Ball is selected + Bag B is selected & Red Ball is selected ie. (1/2)*(4/9) + (1/2)*(5/11) =  (1/2)[ (44+45) / 99 ] = 89/198 Crack CAT with Unacademy! Use referral code HANDA to get 10% off. Daily Live Classes Live Tests and Quizzes Structured Courses Personalized Coaching

# Solved Example #3

February 9th, 2013 by

Question : Find the last two digits of (24)^37298    Answer Last two digits of 24^Odd = 24 Last two digits of 24^Even = 76 Crack CAT with Unacademy! Use referral code HANDA to get 10% off. Daily Live Classes Live Tests and Quizzes Structured Courses Personalized Coaching Enroll Now Download Unacademy App

# Solved Example #2

February 9th, 2013 by

Question: A fair coin is tossed 43 times. What is the number of cases where number of ‘Head’ > number of ‘Tail’? (A) 2^43 (B) (2^43)-43 (C) 2^42 (D) None of the above.   Answer: No. of heads > No. of tails => No. of heads can be 22, 23, 24... 43 Therefore  43C22 + 43C23 + 43C24 .. 43C43 We know that 43C0 + 43C1 + 43C2 ... 43C43 = 2^43 We also know know that our answer has 22 terms which are 43C22 + 43C23 + 43C24 .. 43C43   These 22 terms are equal to = 43C0 + 43C1 + 43C2 .. 43C21 Both of them are equal and add up to 2^43