In a class, 30% of the students offered English, 20% offered Hindi and 10% offered both. If a student is selected at random, what is the probability that he. has offered English or Hindi ?

Correct Answer: 1/6

Explanation:

P(first letter is not vowel) =  2 4

 

P(second letter is not vowel) =  1 3

 

So, probability that none of letters would be vowels is =  2 4 × 1 3 = 1 6

Correct Answer: 15/39

Explanation:

P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize) 

 

 =  1 - 16 C 1 × 14 C 1 × 12 C 1 × 10 C 1 16 C 4 = 15 39

Correct Answer: 3/13

Explanation:

Clearly, there are 52 cards, out of which there are 12 face cards.

P (getting a face card) = 12/52=3/13.

Correct Answer: 1/2

Explanation:

Here S= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}.

Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}.

P(E) = n(E) / n(S)

      = 4/8= 1/2

Correct Answer: 4/7

Explanation:

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P (drawing a white ball) = 8 /14 = 4/7.

Correct Answer: 2/3

Explanation:

Possible outcomes = (RY, YP, PR)

2C1 4C1 + 4C1 6C1 + 6C1 2C1

Required probability = (2C1 4C1 + 4C1 6C1 + 6C1 2C1)/(12C2)

= 8 + 24 + 12/66

= 44/66

= 2/3.

Correct Answer: 7/8

Explanation:

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) =n(E)/n(S)=7/8.

Correct Answer: 551/15134

Explanation:

Total number of elementary events = 50 C 5
Given,third ticket =30

 

 

 

=> first and second should come from tickets numbered 1 to 29 = 29 C 2 ways and remaining two in 20 C 2 ways.

 

 

 

Therfore,favourable number of events =  29 C 2 * 20 C 2

 

 

 

Hence,required probability =  29 C 2 * 20 C 2 / 50 C 5 =551 / 15134

Correct Answer: 35/100

Explanation:

Let   A = Event that A speaks the truth

B = Event that B speaks the truth 

Then P(A) = 75/100 = 3/4

P(B) = 80/100 = 4/5

P(A-lie) =  1 - 3 4= 1/4 

P(B-lie) = 1 - 4 5= 1/5

 

Now, A and B contradict each other =[A lies and B true] or [B true and B lies]

 = P(A).P(B-lie) + P(A-lie).P(B) 

 =  3 5 * 1 5 + 1 4 * 4 5 = 7 20  

 =  7 20 * 100= 35%

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