Explore topic-wise InterviewSolutions in .

49%

Let A be the EVENT that a boy wins 1ST ROUND
Let B be the event that a girl wins 2nd round.
Then, A' = Event that the boy losses 1st round and B' = event that the girl losses 2nd round.
Therefore, P(A) = 55/100 = 11/20, P(B) = 60/100 = 12/20.
P(A') = 1- (11/20) = 9/20 and P(B') = 1- (12/20) = 8/20
First, we have to find the probability that they contradict each other;
That is, P( A And B contradicts each other)
= P[(boy WIN in 1st round And girl losses in 2nd round) (Or) (boy losses in 1st round And girl wins in 2nd round)
= P[(A And B') Or (A' And B)] = P[(A And B')] + p[(A' And B)]
= P(A) x P(B') + P(A') x P(B)
= 11/20 x 8/20 + 9/20 x 12/20
= 88/400 + 108/400
= 196/400
We have to find the %.
Required % = (196/400)x100 = 49%. 

HYPERGEOMETRIC DISTRIBUTION

Hypergeometric distribution

<P> P(X‰¤x)

P(X≤x)

1/2

In a simultaneous throw of two dice, we have n(s) = 6 x 6 = 36
Let E = EVENT of getting two numbers WHOSE sum is odd.
Then E = { (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)}
therefore, n(E) = 18
And P(E) = p( getting two numbers whose sum is odd)
P(E) = n(E)/n(s) = 18/36 = 1/2
Hence the answer is 1/2 

13/102

NA

<P> P(X < 0)

P(X < 0)

13/42

NA

4/13

Total NUMBER of cases = 52
Total face cards = 16 [FAVOURABLE cases]

So probability = 16/52 = 4/13

3/8

NA

1,1/4

NA

1/15

NA

25/39

NA

all of above

all of above

3/4

NA

1/2

Total CASES = [H,T] - 2
Favourable cases = [T] -1
So PROBABILITY of GETTING TAILS = 1/2

4/7

NA

5/7

NA

2/7

NA

20/63

NA

<P> 5/18

In a simultaneous throw of two dice, we have n(s) = 6 x 6 = 36
Let E = event of getting two numbers whose SUM is at most 5.
Then E = { (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}
therefore, n(E) = 10
And P(E) = p( getting two numbers whose sum is at most 5)
P(E) = n(E)/n(s) = 10/36 = 5/18
Hence the ANSWER is 5/18 

3/4

NA

4/7

Let S be the SAMPLE space.
Total number of students in the CLASS = 12 boys + 16 girls = 28.
Then, n(S) = 28.
Let E be the event of CALLING one of them by ENROLL number.
Given that, number of girls = 16.
Then, n(E) = 16.
The PROBABILITY that the one called is a girl = n(S)/n(E) = 16/28 = 4/7. 

2/5

NA

9/20

NA

1/2

NA

X - ¼

x - μ ⁄ σ

17/20

NA

1/m!

NA

We know that if X is a RANDOM variable under binomial distribution, then

Variance of X is given by

Var(X) = NP(1-p)

Thus, in the given question

Var(X) = 10 * 0.3(1-0.3)

Var(X) = 3 * 0.7

Var (X) = 2.1

NON NEGATIVE

Non Negative

5/9

NA

13/102

NA

3/4

Sample space (All [POSSIBLE outcomes)S=(HH,HT,TH,TT)
Event(required outcomes)=(TT,HT,TH)
P(E)=N(E)/n(S)=3/4

10/21

NA

11/15

NA

26/52=1/2

There are 13 CARDS of club and 13 card of DIAMOND. Addition rule is USED here

1/5

NA

3/13

NA

7/15

NA

0.6976

NA

1/2

NA

0.49

NA

10/21

Possible out comes(E) or Favourable OUTCOMES are
E1.2 red ball or
E2.2 green ball or
E3.1 red and 1 green
Number of selecting 2 BALLS from two red balls n(E)=2C2=1
Number of ways of selecting 2 balls from 3 green balls n(E2)=3C2=3
Number of ways of selecting 1 red ball and 1 green ball n(E3)= 2C1*3C1=2*3=6
Sample SPACE S=Number of ways of selecting 2 balls from 7(2+3+2) balls n(S)=7C2=21
n(E)=n(E1)+ n(E2)+ n(E3)=1+3+6=10
Required probability P(E)=n(E)/n(S)=10/21

1/2

NA

33/91

NA

Probability distribution

Probability distribution is listing of the all POSSIBLE OUTCOMES from an experiment with their ASSOCIATED probability for each OUTCOME.

2/9

NA

1/26

Total NUMBER of CASES = 52
Favourable cases = 2

Probability = 2/56 = 1/26

ARITHMETIC MEAN (AM)

Arithmetic Mean (AM)

18/35

NA