Chap-1.pptx statistics chapter one for BBA

na

1. What is probability? The chance that something will happen is called probability. Probability = If A is an event, probability of occurring event A is: P =  

2. What is experiment/trail? The activity that results in, or produces, an event is regarded as experiment. It is not possible to predict exactly the outcomes of the experiment. Example: Tossing a coin; Two events {Head or Tail} Rolling a dice. Six events {1,2,3,4,5,6}

3. What is event? One or more of the possible outcomes of doing something, or one of the possible outcomes from conducting an experiments. Example: A box contains 8 red, 3 blue and 9 green balls. If 3 ball drawn at random determine the probability that: All 3 are red; All 3 are blue. Here, we can let the event of red ball be : A and event of blue ball be : B

4. What is mutually exclusive event? Two or more events that cannot happen together simultaneously in a single trail are termed as mutually exclusive event. Example: The numbers 2 cannot occur simultaneously on the roll of a dice. Events A, B and C are said to be mutually exclusive if A ∩B∩C = Ø

5. What is conditional probability? The probability of one event occurring is that another event has occurred. Example: In a class of 40 BBA students, 20 play cricket, 25 play football and 10 play both. If a player is selected random, find the probability of playing football if a student plays cricket. P(A/B)=

6. Problems on Cards

7. Problem: 1 A card is drawn from a pack of cards. Calculate the probability of getting: King or red card Q ueen and black Black queen and club Black queen or black jack Red king and red

8. Solution:1 We know, Total cards = 52 Red cards = 26 Black cards =26 Club = 13 King = 4 Queen = 4 Jack = 4

9. Let the event of getting a king or red card be: A So, P(A) = = = = = Ans.  

10. Let the event of getting a queen and black card be: A So, P(A) = = = = Ans.  

11. Let the event of getting a black queen and club be: A So, P(A) = = = ans.  

12. Let the event of getting black queen or black jack be: A So, P(A) = = = = = Ans.  

13. Let the event of getting red king and red card be: A So, P(A) = = = = Ans.  

14. Problem : 2 3 card are drawn from a pack of cards. Calculate the following probabilities: 1 queen or 1 ace 2 queens and 1 red At best 1 king At least 1 ace 2 kings 2 hearts

15. Solution:2 We know, Total cards = 52 Red cards = 26 Hearts = 13 King = 4 Queen = 4 Ace = 4

16. Let the event of getting 1 queen or 1 ace be: A So, P(A) = = P(1Q&2others) or P(1ace& 2 others) = + = + = Ans.  

17. Let the event of getting 2 queens and 1 red be: A So, P(A) = = P(2redQ &1red)+P(1redQ &1black Q and 1 red) + P(2 blackQ&1red) = + + = = Ans.  

18. Let the event of getting at best 1 king be: A So, P(A) = = P(1 king &2 others)+P(3 others) = + = Ans.  

19. Let the event of getting at least 1 ace be: A So, P(A) = =P(1ace&2others) or P(2ace&1others) or P(3ace) = + = Ans.  

20. Let the event of getting 2 kings be: A So, P(A) = = P(2 kings & 1others) = = Ans.  

21. Let the event of getting 2 hearts be: A So, P(A) = = P(2 hearts & 1others) = = Ans.  

22. Problems on Balls

23. Problem: 3 A box contains 2 green, 3 black, and 5 yellow balls. If 2 balls are drawn at random, what will be the probability of getting: 2 yellow 1 black 1 yellow and 1 green B alls are same color Balls are different color

24. Solution:3 Given that, Green balls = 2 Black balls = 3 Yellow balls = 5 Total balls = 2+3+5 = 10 So, the probability of getting 2 balls from 10 balls = 10 c 2 = 45

25. Let the event of getting 2 yellow balls be: A So, P(A) = = = ans.  

26. Let the event of getting 1 black ball be: A So, P(A) = = P(1 black & 1others) = = ans.  

27. Let the event of getting 1 yellow and 1 green ball be: A So, P(A) = = P(1 yellow & 1green) = = ans.  

28. Let the event of getting the same color balls be: A So, P(A) = = P(2 green) or P(2 black) or P(2 yellow) = + + = ans.  

29. Let the event of getting the different color balls be: A So, P(A) = = P(1 green & 1 black ) or P(1 green & yellow) or P(1 black & 1 yellow) = + + = ans.  

30. Problem: 4 A box contains 5 red, 7 black, and 8 green balls. If 2 balls are drawn at random: without replacement with replacement Find the probability that the balls will be black or green.

31. Solution:4 Given that, Red balls = 5 Black balls = 7 Green balls = 8 Total balls = 5+7+8 = 20

32. Without replacement Let the event of getting black or green balls be: A So, P(A) = = P(1 st black & 2 nd black) or P(1 st green and 2 nd green) = + = ans.  

33. With replacement Let the event of getting black or green balls be: A So, P(A) = = P(1 st black & 2 nd black) or P(1 st green and 2 nd green) = + = ans.  

34. What Is Bayes' Theorem? Bayes' theorem, named after 18th-century British mathematician Thomas Bayes, is a mathematical formula for determining conditional probability. Conditional probability is the likelihood of an outcome occurring, based on a previous outcome occurring. Bayes' theorem provides a way to revise existing predictions or theories (update probabilities) given new or additional evidence. In Marketing, Bayes' theorem can be used to rate the risk of lending money to potential borrowers.

35. Problem: 5 A manufacturing firm produces steel pipes in three plants with a daily production volumes of 500, 1000, and 2000 units respectively. From past experience, it is known that the fractions of defective output produced by three plants are respectively at random 0.005, 0.008, and 0.010. If a pipe is selected from a day’s total production and found to be defective, find out- From which plant for this defective pipe, the probability is highest. What is the probability that it came from the first plant?

36. Solution:5 Let, B 1 = Production volume of 1 st plant B 2 = Production volume of 2 nd plant B 3 = Production volume of 3 rd plant a nd A = a defective pipe Given that, P(B 1 )= P(B 2 )= P(B 3 )= P( = P( = P( =  

37. Using Bay’s Theorem, The probability of the defective pipe produced from 1 st plant : P(B 1 /A)= = = P(B 1 /A ) = 0.0806  

38. The probability of the defective pipe produced from 2 nd plant : P(B 2 /A )= = = P(B 2 /A) = 0.2627  

39. The probability of the defective pipe produced from 3 rd plant : P(B 3 /A )= = = P(B 3 /A) = 0.6567  

40. As P(B 3 /A) has the highest probability, so we can say that it is most likely to be the defective item has been drawn from the 3 rd plant. Ans. The probability that it came from the first plant is given by P(B 1 /A ) = 0.0806 Ans.