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All posts related to Probability

Nupur
Shared from P & C and Probability - I on Oct 03, 2015 6:18 PM

Numbers of four different digits are formed by using the digits 1, 2, 3, 4, 5, 6 and 7 in all possible ways without repetition. How many of them are greater than 3400?

 
840
560
480
120
None of these
Please type your answer before submitting.

Right ans is 2

Ankush Sharma
Shared from Probability - 4 on Oct 03, 2015 6:16 PM

In a simultaneous throw of two dice, the probability of getting a total of 7 is

 
1 7
7 36
1 4
1 6
Please type your answer before submitting.

Total no. of chance of getting sum 7 will be equal to 6..therefore and will be option d

Abhishek Anand
Shared from Probability on Sep 06, 2015 9:12 PM

Two cards are drawn at random from a pack of 52 cards. The probability that one will be black and the other will be red is

 
None of these
Please type your answer before submitting.

1/2

Himanshu Thapliyal
Shared from IBPS PO (Prelims) Mock - 3 on Aug 31, 2015 1:29 AM

If two squares are chosen at random on a chess board, the probability that they have a side in common is

 
None of these
Please type your answer before submitting.

Two squares are chosen at random on a chessboard. What is the probability that they have a side in common? I think the answer is 1/9. In a chessboard the 4 squares at corners share a side only with 2 squares. If the first square is at corner, then the other square must be from those 2. out of four one square can be chosen in 4c1 ways; hence total no. of ways = 4*2=8 the remaining 24 squares of sidemost rows share a side with 3 squares; hence total no. of ways = 24*3=72 other 36 squares share a side with 4 squares; thats y total no. of ways= 36*4=144 144 72 8=224 are the no. of ways in which 2 squares sharin a side can be selected. total no of ways of selecting 2 squares out of 64 squares are= 64c2 the required probability= 224/(64c2)= 1/9

Himanshu Thapliyal
Shared from IBPS PO (Prelims) Mock - 1 on Aug 29, 2015 1:03 AM

If three sticks are selected at random from a bag containing 9 sticks of length 2, 3, 4, 5, 6, 7, 8, 9 and 10 units, then what is the probability that a triangle can be formed with those three sticks?

 
None of these
Please type your answer before submitting.

Triangle inequality theorem states that sum of the lengths of any two sides of a triangle must be greater than the third side.

Suppose a,b,c are the sides of a triangle. Then
a + b > c
a + c > b
b + c > a

But we don't need to check all these conditions. Just add the shorter sides and compare with the longer side.
If sum of the shorter sides is greater than the longer side, triangle can be formed. Else, triangle cannot be formed.

With this information, we can approach the given problem.

Total ways in which 3 numbers can be selected from (2, 3, 4, 5, 6, 7, 8, 9,10) = 9C3 = 84

Using the triangle inequality theorem, we can see that the following combinations cannot form a triangle.

(2,3,5), (2,3,6), (2,3,7), (2,3,8), (2,3,9), (2,3,10)
(2,4,6), (2,4,7), (2,4,8), (2,4,9), (2,4,10)
(2,5,7), (2,5,8), (2,5,9), (2,5,10)
(2,6,8), (2,6,9), (2,6,10)
(2,7,9), (2,7,10)
(2,8,10)

(3,4,7), (3,4,8), (3,4,9), (3,4,10),
(3,5,8), (3,5,9), (3,5,10),
(3,6,9), (3,6,10)
(3,7,10)

(4,5,9),(4,5,10)
(4,6,10)

i.e., 34 combinations cannot form triangle out of the total 84

P(selected sides cannot form a triangle) =34/84

P(selected sides can form a triangle) = 1−34/84=50/84=25/42