Probability and Statistics

Probability:

1) A bag contains 8 white and 6 black balls. 3 balls are drawn at random. Find the probability that they are of same color.

Explanation: Let S be the sample space

Then n(S) = no of ways of drawing 3 balls out of (8+6)

=14C3 

=14! / 3! * 11!

= 364

Let E = event of getting both balls of same colour

Then, n(E) = no of ways (3 balls out of 8) or (3 balls out of 6)

                   = 8C3 + 6C3  = 56 + 20 = 76

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

                         = 76 / 364

2) Two cards are drawn at random from a pack of 52 cards. what is the probability that either both are black or both are queen?

Explanation: We have n(s) =52C2  = 52*51/2*1= 1326.

Let A = event of getting both black cards

B = event of getting both queens

A∩B = event of getting queen of black cards

n(A) = (52*51)/(2*1) = 26C2 = 325,

n(B)=  (26*25)/(2*1)= 4*3/2*1= 6  and  n(A∩B) = 4C2 = 1

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and

P(A∩B) = n(A∩B)/n(S) = 1/1326

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

3) 3 dice are tossed. The probability that the total score is a prime number is:

Explanation: Clearly, n(S) = (6 x 6 * 6) = 216.

Let E = Event that the sum is a prime number.

Then E= { (1, 1,1),  . . .}

n(E) = 15.

P(E) = n(E)/n(S) = n(E) / 216 = 

 4) A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are blue, is:

Explanation: Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15

= 15C3  =(15*14*13)/(3*2*1)= 455.

Let E = event of getting all the 3 blue balls.

n(E) = 6C3 =  20.

=> P(E) = n(E)/n(S) = 10/455 = 2/91.

5) Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

Explanation: Let S be the sample space.

Then, n(S) = 52C2=(52 x 51)/(2 x 1) = 1326.

Let E = event of getting 1 spade and 1 heart.

n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = (13C1)*(13C1) = 169.

P(E) = n(E)/n(S) = 169/1326 = 13/102.

6) What is the probability of getting 53 Mondays in a leap year?

Explanation: 1 year = 365 days . A leap year has 366 days

A year has 52 weeks. Hence there will be 52 Sundays for sure.

52 weeks = 52 x 7 = 364days

366 – 364 = 2 days

In a leap year there will be 52 Sundays and 2 days will be left.

 

These 2 days can be:

1. Sunday, Monday

2. Monday, Tuesday

3. Tuesday, Wednesday

4. Wednesday, Thursday

5. Thursday, Friday

6. Friday, Saturday

7. Saturday, Sunday

Of these total 7 outcomes, the favourable outcomes are 2.

Hence the probability of getting 53 days = 2/7

7) 3 dice are thrown simultaneously. Find the probability that all of them show the same face.

Explanation:

Total events we get throwing 3 dice simultaneously is:

n(s)= 6 * 6 * 6 = 216

Getting same Face:

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

n(X) =6

P(E)=n(X)/ n(s) = 6/216 =1/36

8) Three unbiased coins are tossed. What is the probability of getting at most two heads?

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, HHH}.

n(E) = = {TTT, TTH, THT, HTT, THH, HTH, HHT} =7

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

9) A box contains 20 bulbs, of which just 6 are defective. If a random sample of 10 bulbs is drawn, find the probability that the sample contains exactly 2 defective bulbs.

Explanation: Total number of elementary events = 20C10

Number of ways of selecting exactly 2 defective bulb out of 6 and 8 non-defective out of 14 is 3C1*7C4

So, required probability =6C2*14C8/20C10 

                                        = 

Types of Statistics

 Descriptive Statistics:  

Statistics dealing with numbers (numerical facts, figures, or information) to describe any phenomena.  These numbers are descriptive statistics.

 e.g.  cricket batting averages, government deficits, Movie Ratings etc.

 

 Inferential statistics

  Inferential statistics is a decision, estimate, prediction, or generalization about a population, based     on sample.

 A population is a collection of all possible individual, objects, or measurements of interest. 

 A sample is a portion, or part, of the population of interest.

Types of Data:

1) Categorical:  Represents categories.

2) Numerical: a) Disctrete

                        b) Continuous

Levels of measurement:

1) Qualitative:  a) Nominal

                         b) Ordinal

2) Quantitative:  a) Interval

                          b) Ratio     

Measures of Central Tendency

  Mean:  mean is the arithmetic average  

Median:  middle value. 

Mode: Most frequently occurring value in data set