A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

Correct Answer: 55/221

Explanation:

n(S) =  C 2 52 = 1326

 

 Let  A = event of getting both red cards

 

and B = event of getting both queens

 

then  A ? B = event of getting two red queens

 

n(A) =  C 2 26 = 325,   n(B) =  C 2 4 = 6

 

  n ( A ? B ) = C 2 2 = 1

 

  ? P A = 325 1326 , P B = 6 1326

 

  P A ? B = 1 1326

 

?P ( both red or both queens) =  P A ? B

 

=  P A + P B - P A ? B = 325 1326 + 1 221 - 1 1326= 55 221

Correct Answer: 2/3

Explanation:

S = { 1, 2, 3, 4, .....18 } 

=> n(S) = 18

 

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

 

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

 

  E 3 = E 1 ? E 2 = { 6 , 12 , 18 } 

=> n(E3) = 3

 

? E = E 1 ? E 2 = E 1 + E 2 - E 3 

=> n(E) = 9 + 6 - 3 =12

 where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

 

? P ( E ) = n ( E ) n ( S ) = 12 18 = 2 3

Correct Answer: 1/5

Explanation:

Number of applicants = 5
On a day, only 1 leave is approved.
Now favourable events =  1 of 5 applicants is approved
Probability that Laxmi priya's leave is granted = 1/5.

Correct Answer: 3/10

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

Correct Answer: 3/4

Explanation:

Let S be the sample space.
Here n(S)= 2 3 = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.

Correct Answer: p/(p+q)

Explanation:

Total number of cases=p+q

Favourable cases=p

Probability of that event is=p/(p+q)

Correct Answer: 1/5

Correct Answer: 9/70

Explanation:

Let A be the event of driver A drive safely after consuming liquor. 

 

Let B be the event of driver B drive safely after consuming liquor. 

 

Let C be the event of driver C drive safely after consuming liquor. 

 

P(A)=2/5,P(B)=3/7,P(C)=3/4

 

The events A, B and C are independent . Therefore,

 

P A ? B ? C = P ( A ) P ( B ) P ( C ) = 2 5 * 3 7 * 3 4 = 9 70

 

Therefore, The probability that all the drivers will drive home safely after consuming liquor is 9/70

Correct Answer: 4/13

Correct Answer: [1, 1/2]

Explanation:

Probability of atleast one failure
= 1 - no failure > 31/32
= 1 - P 5 > 31/32
=   P 5 <1/32
= p < 1/2
Also p > 0
Hence p lies in [0,1/2].