Quiz-1

Total no. of balls 12
n(S)= 12C3=220
3balls can be selected from 3red+4green
=7 balls in 7C3 ways
7C3=35
probability 35/220=7/44

Let us assume the 3 girls as a single unit.

So we've now 4 boys +1 = 5 units which can arrange themselves in 5! = 120 ways

. Now in these, 3 girls can arrange themselves in 3!= 6 ways.

So finally we have the favorable cases= 120*6=720

The sample space consists of all possible arrangements of 4 boys and 3 girls which is = (4+3)!= 7! = 5040

Required probability = (favourable cases)/(sampl space)=720/5040 = 72/504=  1/7= 2/14

Any positive integer can be represented as;

7n+1

7n+2

7n+3

7n+4

7n+5

7n+6

When the integer is divided by 7, it leaves the remainder 1 Therefore the integer has to be of type 7n+1 Therefore the probability of choosing a number of the type 7n+1 is 1/7

We use probability trees in this case. The probability that John will be appointed and Ganguly appointed is : 0.5 and 0.7

This is : 0.5 × 0.7 = 0.35

The probability that Greg will be appointed and Ganguly be appointed is : 0.3 and 0.6

This is: 0.3 × 0.6 = 0.18

The probability that Gary will be appointed and Ganguly be appointed is : 0.2 and 0.5

This is : 0.2 × 0.5 = 0.1

The probability of Ganguly appointed is : 0.35 or 0.18 or 0.1

This is : 0.35 + 0.18 + 0.1 = 0.63

possible cases are (1,1) (1,2) and (2,1) = 3

3/36 = 1/12

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*4/2*1 = 10.

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

P( None is defective)

= 16C2 / 20C2

= (16x15/2x1 ×2x1/20x19) = 12/19.

P( at least one is defective) = (1- 12/19) = 7/19.

Let p(A) = 30% = 0.3  P(B) = 20% = 0.2

Both, p(AnB) = 10% = 0.1

So, p(AuB) = p(A)+p(B)-p(AnB) P(AuB)

= 0.3+0.2-0.1 = 0.4 = 4/10

= 2/5

All are different size ball so no two are of the same size and only one ball among each color of the ball is smallest so in the case of the smallest red ball

A total red ball

It is given that there are 6 red balls, 7 green balls and 5 blue balls. Therefore, the total number of balls = 18.

P(red ball) = 6/18.

P(smallest ball and red ball) = 1/18

Because out of all the 18 balls, only one is the smallest red.

Therefore, P(The smallest red ball is picked | a red ball is picked) = P(smallest ball and red ball)/P(red ball)

\frac{\frac{1}{18}}{\frac{6}{18}}

=1/6

 

P(A/B) = P(A ∪ B)/P(B)

P(A / B) = P(A)P(B)/P(B) = 0:2/0.4 =0.5