CoCubes Probability Question and Answers Quiz-1

Total balls = 12
No. of balls except yellow = 12-5=7
probability of first ball as not yellow = 7/12

probability of second non yellow ball when first non yellow ball is already drawn = 6/11

similarly, probability of third non yellow ball = 5/10
Probability of three non yellow balls drawn randomly = (7/12)*(6/11)*(5/10) = 210/1320 = 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've 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 = (favorable cases)/(sample-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. Probability that Sachin will be appointed and Virat appointed is : 0.5 and 0.7

This is : 0.5 × 0.7 = 0.35

The probability that will be appointed and Virat be appointed is : 0.3 and 0.6 This is: 0.3 × 0.6 = 0.18

The probability that Ashwin will be appointed and Virat be appointed is : 0.2 and 0.5 This is : 0.2 × 0.5 = 0.1

The probability of Virat appointed is : 0.35 or 0.18 or 0.1 This is : 0.35 + 0.18 + 0.1 = 0.63

Two dice are rolled together

Hence the sample space, n(S) = 6 \times 6 = 36

Let A be the probability that sum of the numbers appearing on the top faces of the dice is less than 4, hence

A = [(1,1) , (1,2) , (2,1)}

n(A) = 3

P(A) =\frac{ (n(A))}{(n(S))} = \frac{3}{36} = \frac{1}{12}

The probability that sum of the numbers appearing on the top faces of the dice is less than 4 is 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 diffrent size ball so no two are of same size and only one ball among the each color of ball is smallest

so  in the case of smallest red ball .

total red ball is 6 and only one ball is smallest so answer is  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