HSBC Probability and Permutation and Combination Questions and Answers Quiz-1

Explanation:

Let S be the sample space

 

Then n(S) = no of ways of drawing 2 balls out of (6+4) =10C2

10 =10*9/2*1

=45

 

Let E = event of getting both balls of same colour

 

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

 

=6C2+4C2

= 6*5/2*1+4*3/2*1

= 15+6 = 21

 

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

Explanation:

We have n(s) =52C2

52 = 52*51/2*1= 1326.

Let A = event of getting both black cards

B = event of getting both queens

A∩B = event of getting queen of black cards

n(A) = 52*51/2*1

= 26C2 = 325, n(B)= 26*252*1= 4*3/2*1= 6  and  n(A∩B) = 4C2

= 1

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and

P(A∩B) = n(A∩B)/n(S) = 1/1326

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

Explanation:

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }

n(E) = 15.

P(E) = n(E)/n(S) = 15/36 = 5/12.

Explanation:

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 = 15C3

=15*14*13/3*2*1

= 455.

Let E = event of getting all the 3 red balls.

n(E) = 5C3

= 5*42*1

= 10.

=> P(E) = n(E)/n(S) = 10/455 = 2/91.

xplanation:

Total number of outcomes possible, n(S) = 10 + 25 = 35

 

Total number of prizes, n(E) = 10

P(E)=n(E)n(S)=10/35=2/7

No. of Digits = 4 All are distinct; They can be arranged in 4! = 24 ways
Each of the digits 3, 4, 5 and 6 occur at unit place = 3! Ways = 6 ways.
Thus there will be 6 numbers ending with 3, 4, 5 and 6 each. So the sum of the digits at unit's place = 6(3 + 4 + 5 + 6) =108
The sum of numbers = 108 × 10³ + 108 × 10² + 108 × 10¹ + 108 × 10⁰ = 119988

Here each of the digits 2 and 3 will occur at unit, tens, hundred and thousand place ( ³P₃/2!) = 3 times. Digit 4 will occur at each place = 6 times;
∴ Sum of digits at unit, tens, hundred and thousand place = 3 × 3 + 6 × 4 + 3 × 2 =39. Sum of numbers formed =
= 39 × 10³ + 39× 10² + 39 × 10¹ + 39 × 10⁰ = 43329

The required number of lines= ¹²C₂ – ⁸C₂ + 1 = 1 + 66 – 28 = 39