CAT4MBA

Q6. If the probability of X failing in an examination is 1/5 and that of Y failing the same examination is 3/10, then the probability that either X or Y fails is :
a. 1/2
b. 11/25
c. 19/50
d. None of these

here both the events are independent to each other

So P(A U B) = P(A) + P(B) - P(A).P(B)

= 1/5 + 3/10 - 3/50

= 1/2 - 3/50

= 22/50

Q4. 11 books consisting 5 management books, two accountant books and four science books are arranged on shelf at random. Find the probability that books of each kind are together?
a. 1/1155
b. 2/1155
c. 3/1155
d. 4/1155

ans:

(5! * 2! * 4! * 3! )/ 11!

= 2! * 4! / 7*8*9*10*11

= 2!/7*3*10*11

= 1/21*55 = 1155

Q3. There are 5 envelopes corresponding to 5 letters. If the letters are placed in the envelopes at random, find the probability that all the letters are placed in wrong envelopes.
a. 119/120
b. 4^5/5^5
c. 20/120
d. 59/60

ans:

(120-1)/120

= 119/120

Q2. The probability of two events E and F are ¼ and ½ respectively. The probability of their simultaneously occurrence is 0.14. Find the probability of occurrence of either E or F?
a. 0.41
b. 0.51
c. 0.61
d. 0.71

ans:

1/4 + 1/2 - 0.14

= 0.75 - 0.14

= 0.61

Q1.  From a bunch of 5 keys, 3 right keys have to be used to open a combination lock, such that the first key opens the lock and then allows the second key to be used , and the second key then allows the third to be used. What are the chances of opening the lock in a random selection of 3 keys?
a. 1/30
b. 1/60
c. 1/120
d. 1/80

selecting 3 key from 5 ,5c3 and then combination of 3 key=3!

 so total ways=60

in only one case we can open lock so probability=1/60

so ans is (b)

Little Star