A box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains exactly one defective bulb.

Correct Answer: 0.7

Explanation:

P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7

Correct Answer: 1/9

Explanation:

One person can select one house out of 3= 3 C 1 ways =3.

 

Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9.

 

Therefore, probability that all thre apply for the same house is 1/9

Correct Answer: 17/42

Explanation:

3 letters can be choosen out of 9 letters in 9 C 3 ways.

 

More than one vowels ( 2 vowels + 1 consonant  or  3 vowels ) can be choosen in  ( 4 C 2 * 5 C 1 ) + 4 C 3 ways

 

Hence,required probability = 4 C 2 * 5 C 1 + 4 C 3 9 C 3 = 17/42

Correct Answer: 5/26

Explanation:

We know that,

Total number of balls n(S) = 26

Number of vowels n(E) = 5

Hence, required probability = n(E)/n(S) = 5/26.

Correct Answer: 19/42

Explanation:

A red ball can be drawn in two mutually exclusive ways

 (i) Selecting bag I and then drawing a red ball from it.

 

(ii) Selecting bag II and then drawing a red ball from it.

 

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore 

P(E1) = 1/2  and  P(E2) = 1/2

Now,  P A E 1 = Probability of drawing a red ball when the first bag has been selected = 4/7

   P A E 2  = Probability of drawing a red ball when the second bag has been selected = 2/6

 Using the law of total probability, we have 

 P(red ball) = P(A) =  P E 1 × P A E 1 + P E 2 × P A E 2 

 

                          =  1 2 × 4 7 + 1 2 × 2 6 = 19 42

Correct Answer: 3/4

Explanation:

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E= {(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)}

n(E) = 27.

P(E) = n(E)/n(S) = 27/36 = 3/4.

Correct Answer: 1/9

Explanation:

In two throws of a die, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) =n(E)/n(S)=4/36=1/9.

Correct Answer: 5/18

Explanation:

n(S) = 36

 

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

 

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

 

n A = 6 , n B = 5 , n A ? B = 1

 

?Required probability =  P A ? B

 

 =  P A + P B - P A ? B

 

=   6 36 + 5 36 - 1 36 =  5 18

Correct Answer: 1/3

Explanation:

Total number of balls = (8 + 7 + 6) = 21.

Let E = event that the ball drawn is neither red nor green 

            = event that the ball drawn is blue.

n(E) = 7.

P(E) = n(E)/n(S) = 7/21 = 1/3.

Correct Answer: 1/216

Explanation:

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:

 

=  6 * 6 * 6 * 6 = 6 4

 

n(S) =  6 4

 

Let X be the event that all dice show the same face. 

 

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

 

n(X) = 6

 

Hence required probability =  n ( X ) n ( S ) = 6 6 4 = 1 216