GRE QUANTITATIVE: PERMUTATIONS, COMBINATIONS AND PROBABILITY
- 2. This video is designed to help students preparing for the GRE QUANT SECTION Do subscribe to my channel Visit my page https://www.mathmadeeasy.co/about-1 You can contact me for personal classes.
- 3. What are permutations? What are combinations? How do we use them in selection and arrangements? Learn all this and much more in this presentation.
- 4. Number of ways of arranging r objects out of n objects = 𝑛 𝑃𝑟 = 𝑛! 𝑛−𝑟 ! 5 𝑃2 = 5 4 = 20 Number of ways of selecting r out of n objects = 𝑛 𝐶 𝑟 = 𝑛! 𝑟! 𝑛−𝑟 ! 5 𝐶2 = 5(4) 1(2) = 10
- 5. Question 1 6 points lie in a circle. How many quadrilaterals can be formed by joining these points.
- 6. We need to choose 4 out of 6 points since a quadrilateral has 4 vertices. Total number of quadrilaterals =6 𝐶4 = 6 𝑐2 = 6×5 1×2 = 15
- 7. Question 2 Quantity A All the jacks are removed from a pack of cards and then a card is drawn. Find the probability of drawing a red king Quantity B 2 unbiased dice are thrown. Find the probability of obtaining 6 as a sum or product
- 8. Quantity A There are 4 jacks which are removed. There are 48 cards left . We need to draw 1 red king from 4 kings. Probability = 4 𝐶1 48 𝐶1 = 4 48 = 1 12
- 9. Quantity B We need 6 as a sum or product Favourable outcomes are = {(1,5), (1,6),(2,4),(2,3), (3,3),(3,2),(4,2),(5,1),(6,1)} 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 = 9 36 = 1 4 A B 1 12 1 4 Cross multiplying, we get A B 4 12 B is greater
- 11. Question 3 Compare the 2 quantities Quantity A In how many ways can an examiner select a set of 6 true/false conditions Quantity B In how many ways can an examiner enter a set of 5 objective questions, each question having 3 choices
- 12. Option A There are 3 options to the true/false question ie true, false or not answered Since there are 6 questions, total number of ways =36 Option B There are 5 questions, each question having 3 choices Total number of ways = 35 Option A is greater
- 13. Question 4 An urn contains 5 blue, 5 red, 5 green and 5 yellow balls. Find the probability that if 2 balls are drawn at random, both are blue.
- 14. Bl R G Y 5 5 5 5 𝑃robability = 5 𝐶2 20 𝐶2 = 5 65
- 16. Question 5 In how many ways can 5 children be seated in a row so that 2 children never sit together
- 17. Treat the 2 children as one. There are now 5 -2 +1 =4 children These can be arranged in 4! =24 ways Number of arrangements of n objects = n! The 2 children within themselves can be seated in 2 ways Total number of ways = (24)(2)=48
- 18. Question 6 How many 5 digit numbers can be formed using 0,2,3,4 and 5 such that repetition is allowed and the number is divisible by 2 or 5
- 19. The last digit can be 0,2 or 5. The last digit can be filled in 3 ways --- --- ----- ----- ----- The first digit cannot be 0 The first digit can be filled in 4 ways Since there is no restriction, the remaining 3 digits can be filled in 53 𝑤𝑎𝑦𝑠(5 × 5 × 5) Answer = 125(3)(4)=1500 ways
- 20. Question 7 Team A has 5 athletes chosen from 7 people Team B has 6 athletes chosen from 8 people Column A Column B Number of unique teams in A Number of unique teams in B
- 21. 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑖𝑞𝑢𝑒 𝑡𝑒𝑎𝑚𝑠 𝑖𝑛 𝐴 = 7 𝐶5 = 7 × 6 1 × 2 = 21 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑖𝑞𝑢𝑒 𝑡𝑒𝑎𝑚𝑠 𝑖𝑛 𝐵 = 8 𝑐6 = 8 × 7 1 × 2 = 28 Column B is greater
- 23. Question 8 If the odds in favour of an event are 4:5, find the probability that it will occur.
- 24. 𝑜𝑑𝑑𝑠 𝑖𝑛 𝑓𝑎𝑣𝑜𝑢𝑟 𝑜𝑓 𝑎𝑛 𝑒𝑣𝑒𝑛𝑡 ℎ𝑎𝑣𝑖𝑛𝑔 probability p = 𝑝 1−𝑝 𝑜𝑑𝑑𝑠 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑡ℎ𝑒 𝑒𝑣𝑒𝑛𝑡 = 1 − 𝑝 𝑝 𝑝 1 − 𝑝 = 4 5 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 , 𝑤𝑒 𝑔𝑒𝑡 𝑝 = 4 9
- 25. Question 8 The probability that A solves the problem is 1 9 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 B solves it is 5 18. . 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑜𝑛𝑙𝑦 1 𝑝𝑒𝑟𝑠𝑜𝑛 𝑠𝑜𝑙𝑣𝑒𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑙𝑒𝑚
- 26. Required answer = P(A𝐵) + 𝑃 𝐴𝐵 = 1 9 13 18 + 8 9 5 18 = 53 18
- 27. Question 9 Given 8 flags of different colours, how many different signals can be made using 3 flags at a time one below the other
- 28. The first flag can be arranged in 8 ways. Since they are of different colours, the 2nd flag can be arranged in7 ways. The 3rd flag can be arranged in 6 ways. Total number of ways = 8 × 7 × 6 = 336
- 29. Question 10 An urn contains balls numbered 6 to 20. Find the probability that the ball drawn is a multiple of 4 or 5 Multiples of 4 : 8, 12, 16,20 Multiples of 5 : 10,15,20 A- multiple of 4 , 𝑃 𝐴 = 4 15 B: multiple of 5, P(B)= 3 15 𝑃 𝐴 ∩ 𝐵 = 1 15 Multiple of 4 and 5 : 20
- 30. P(A∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) = 4 15 + 3 15 − 1 15 = 6 15 = 2 5
- 31. Question 11 An urn contains 5 red, 6 blue, and 1 green ball. Another urn contains 6 red, 8 blue and 2 green balls. One ball is drawn at random from each urn. Find the probability that both are green.
- 32. R BLUE G 5 6 1 R B G 6 8 2 P(𝐺1 𝐺2) = 𝑃 𝐺1 𝑃(𝐺2)= 1 12 2 16 = 1 96
- 33. What did we learn? How to apply permutations and Combinations to solve problems