Quiz – 1

Let S be the sample space

Then, n(s)= number of ways of drawing 4 balls out of 18

=18C4

= (18x17x16x15)/ (4x3x2x1)

=3060

Let E = event of getting all the 4 yellow balls.

n(E)= 4C4= 1

P(E)=  n(E) / n(S)

P(E)=  1/3060

Since Maximum portability is 1.
We can get total probability of non defective bulbs and subtract it form 1 to get
total probability of defective bulbs.
Total cases of non-defective bulbs
=>20C3= (20x19x18)/ (3x2x1)=1140
Total Cases
=>25C3= (25x24x23)/ (3x2x1)=2300
Probability= 1140/2300=57/115
Probability of atleast one defective = 1- (57/115)
=58/115

P (E) = 40/100= 2/5
P (H) = 30/100= 3/10
P (E Π H) = 5/100= 1/20
P (E U H) = P(E) + P (H) - P (E Π H)
=2/5 + 3/10 – 1/20
= (8+6-1)/20
= 13/20

Pr (The smallest green ball is picked |a green ball is picked)

= Pr (The smallest green ball is picked & a green ball is picked) / Pr (a green ball is picked)

= Pr (The smallest green ball is picked & a green ball is picked)= 1/14

= Pr (a green ball is picked) = 3/14

Therefore:

Pr (The smallest green ball is picked| a green ball is picked) = (1/14 )/(3/14)

= 1/3

P(S/T) = P(S ∪ T)/ P(T)
P(S ∪ T) = P(S)P(T)
P(S/T)= (0.7*0.8)/0.8 =0.7

If a number is divisible by 5 is formed using the digits 5, 6, 7, and 0, the number must have two digits at the last. These can be:

05, 50, 60, 65, 55, 70, 75.

In each of these cases, the four digit number can be formed using the remaining 2 digits in 2! = 2 ways.

A number divisible by 5 can be formed in 2 X5 = 10 ways

Total number that can be formed using the digits 5, 6, 7, and 0 and without 5 repeated is 4! = 24

required probability= 24/10 = 12/5

: A dice has six faces with numbers 1, 2, 3, 4, 5, and 6 one on each side.

When an unbiased dice is thrown, the probability of any of the faces to appear in a single role is 1/6.

Or,

P(1)=P(2)=P(3)=P(4)=P(5)=P(6)=1/6

Now that number is greater than 5 and 6

So it can be P(5) or P(6)=

P(5)+P(6)=2/6= 1/3

Total balls= 9
Chances of selecting a Red or a Blue ball (4 +2)= 6
Hence the probability is 6/9 = 2/3.

Total number of sample space two pencils can be drawn = 7^C^2 = 21

At least one to be Red= 3^C^1*4^C*1 + 3^C^2*4^C^0= 15

Probability = 21/15

Total Apples in bags M and N = 9

Total number of fruits in both the bags= (7 + 11)=  18

Probability that both the fruits are Apples: 9/18= ½.

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