The empirical relation between Mean, Median and Mode is

Mode + 2Mean = 3 Median

⇒ Mode = 3 × 22 – 2 × 21

= 66 – 42 = 24.

In a single throw of two dice, find

P (an odd number on the first die and a 6 on the second)

Total outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) ,

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) ,

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Desired outcomes are (1, 6), (3, 6), (5, 6)

Total no.of outcomes are 36 and desired outcomes are 3

Therefore, probability of getting odd on the first die and 6 on the second die

=3/36=1/12

Conclusion: Probability of getting odd on the first die and 6 on the second die, when two dice are rolled is 1/12

We know that,

Probability of occurrence of an event

By permutation and combination, total no.of ways to pick r objects from given n objects

is nCr

Now, total no.of ways to pick a ball from 20 balls is 20C₁ = 20

Our desired output is to pick a white or red ball.

So, no.of ways to pick a white or red

ball from 16 balls(because there are a total of 16 balls which are either red or white) is 16C₁ = 16

Therefore, the probability of picking a white or red ball =16/20=4/5

Conclusion: Probability of picking a white or red ball from 9 red, 7 white, and 4 black balls is 4/5.

We know that,

Probability of occurrence of an event

Let T be tails and H be heads Total possible outcomes = TTT, TTH, THT, HTT, THH, HTH, HHT, HHH

Desired outcomes are exactly two tails.

So, desired outputs are TTH, THT, HTT

Total no. of outcomes are 8 and desired outcomes are 3

Therefore, the probability of getting exactly 2 tails =3/8

Conclusion: Probability of getting exactly two tails is 3/8

We know that,

Probability of occurrence of an event

Total no. of outcomes are 12

Desired output is picking a number which is multiple of 2 or 3.

So, desire outputs are 2, 3, 4, 6, 8, 9, 10, 12.

Total no. of desired outputs are 8

Therefore, the probability of getting a number which is multiple of 2 or 3 = 8/12= 2/3

Conclusion: Probability of picking a ticket which is multiple of 2 or 3 is 2/3

We know that,

Probability of occurring = 1 - the probability of not occurring

Let’s calculate for the probability of not occurring, i.e. probability such that both of them don’t have a birthday on the same day. For suppose the first person has a birthday on a particular day then the other person can have a birthday in the remaining 364 days

Probability of not having the same birthday=364/365

Probability of having same birthday = 1 – probability of not having the same Birthday

Conclusion: Probability of two persons having the same birthday is 1/365

Given: P(A) = 0.60, P(A or B) = 0.85 and P(A and B) = 0.42

To find: P(B) Formula used : P(A or B) = P(A) + P(B) - P(A and B)

Substituting in the above formula we get,

0.85 = 0.60 + P(B) – 0.42

0.85 = 0.18 + P(B)

0.85 – 0.18 = P(B)

0.67 = P(B)

P(B) = 0.67

Given: A and B are mutually exclusive events P(not A) = P  = 0.65 , P(A or B) = 0.65

To find: P(B)

Formula used:

P(A) = 1 – P

P(A or B) = P(A) + P(B) - P(A and B)

For mutually exclusive events A and B, P(A and B) = 0

P(A) = 1 – P(not A)

P(A) = 1 – 0.65

P(A) = 0.35

Substituting in the above formula we get,

0.65 = 0.35 + P(B)

P(B) = 0.65 – 0.35

P(B) = 0.30