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Given:

Digits =2, 3, 5, 6, 7 and 9

Concept used:

Divisibility rule of 5:

The number is divided by 5, if theunit digit of a number is either 5 or 0.

Calculation:

Unit digit can only be 5.

There is only 1 possible way to fillunitplace.

Remaining placescan be filled by 2, 3, 6, 7 or9.

There is 5possible ways to fill ten'splace.

There is 4 possible ways to fill hundredth place as digits cannot be repeated.

Total numbers that can be formed = 1× 5× 4= 20

∴ 3 - digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 is 20.

1/6

NA

9/10

NA

999/100

NA

18/35

Let A be the event of the GROUP A pass
Let B be the event of the group B pass
Then, A'= Event of the group A's FAIL and B'= event of the group B's fail.
Therefore, p(A) = 2/7 and p(B) = 2/5,
P(A') = 1 - P(A) = 1- 2/7 = 5/7 and P(B') = 1- P(B) = 1- 1/5 = 4/5
Required PROBABILITY = P[( A And B') Or (B And A')]
= P[( A And B') Or (B And A')]
= P[( A And B') + (B And A')]
= P[( A And B')] + p[(B And A')]
= p(A) X p(B') + P(A') x P(B)
= (2/7 x 4/5) + (2/5 x 5/7) 
= (8/35 + 10/35) = 18/35

833/858

Total number of DRINK bottles = 6 + 3 + 4 = 13.
Let S be the sample space.
Then, n(S) = number of ways of taking 3 drink bottles out of 13.
Therefore, n(S) = 13C3 = (13 x 12 x 11)/(1 x 2 x 3) = 66 x 13 = 858.
Let E be the event of taking 3 bottles of the same variety.
Then, E = event of taking (3 bottles out of 6) or (3 bottles out of 3) or (3 bottles out of 4)
n(E) = 6C3 + 3C3 + 4C3
= 6 x 5 x 4 / 1 x 2 x 3 + 1 + 4 x 3 x 2 / 1 x 2 x 3
= 20 + 1 + 4 = 25.The probability of taking 3 bottles of the same variety = n(E)/n(S) = 25/858.
Then, the probability of taking 3 bottles are not of the same variety = 1 - 25/858 = 833/858.

1/2

NA

1/455

NA

41/44

41/44

3/4

Total 4 cases = [HH, TT, TH, HT]
FAVOURABLE cases = [HH, TH, HT] 
PLEASE note we need atmost one tail, not atleast one tail.

So probability = 3/4

np

np

The SALE or PURCHASE PRICE of a HOUSE

The sale or purchase price of a house

3/4

Total NUMBER of CASES = 6*6 = 36

Favourable cases = [(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)] = 27

So PROBABILITY = 27/36 = 3/4

7/8

NA

 1/(365)2

NA

23/35

NA

9/13

NA

1/16

1/16

0.5904

NA

1/2

NA

12/36

NA

nk/N

nk/N

MUST have at LEAST 3 TRIALS

must have at least 3 trials

55/221

NA

1/2

Total CASES = [H,T] - 2
Favourable cases = [H] -1
So PROBABILITY of GETTING HEAD = 1/2

1/36

NA

C

C

NONE of the above

NA

35%

Probability that A is TELLING the truth P(A)=75/100=3/4
Probability that B is telling the truth P(B)=80/100=4/5
Probability that A is LYING P(A’)=1-75/100 =25/100=1/4
Probability that B is lying P(B’)=1-80/100 =20/100=1/5
Contradiction means either of 

<P> p = Q

p = q

4/13

NA

1/6

In a simultaneous throw of two DICE, we have n(s) = 6X6 = 36
Let E = event of GETTING two numbers are same.
Then E = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
therefore, n(E) = 6
And P(E) = p( getting two numbers are same)
P(E) = n(E)/n(s) = 6/36 = 1/6
Hence the ANSWER is 1/6 

trials

trials

4/5

NA

316/435

NA

7/19

7/19

1

1

5/8

NA

0 and 1

0 and 1

3/4

NA

2

2

2/13

NA

15/64

NA

3/5

Given that, 45% play football; that is, P(F) = 45/100 = 9/20
30% play VOLLEY ball; that is, P(V) = 30/100 = 6/20
And, 15% play both volleyball and football; that is, P(F And V) = 15/100 = 3/20
Now, we have to FIND the probability that 1 STUDENT plays football or volley ball;
 that we have to find, P(F or V)
We know that, P(F Or V) = P(F) + P(V) - P(F And V)
= 9/20 + 6/20 - 3/20 = 12/20 = 3/5.
Hence, the required probability 3/5. 

18/455

NA

0.36

NA

18/35

NA

2/3

Total CASES = 10 + 20 = 30
Favourable cases = 20

So PROBABILITY = 20/30 = 2/3

1/8

NA

HYPERGEOMETRIC DISTRIBUTION

Hypergeometric distribution