Student Number:21448762 STA1DCT Assignment 4 Name: Khalid Hasan Rohan "This is my own work. I have not copied any of it from anyone else." Question 1: a) Answer 54,264 different random samples of size n = 6 are possible from a population of N = 21 01 (a) The number of different random samples of size=6(1) that can be chosen from population N =21. . 01 bar By combination formular Cn = n! malt words of 2 How to reduce of a: (n=7)! 11 21! er = 0 6:(21-6):01 =21×20×19×18x17×16×15! 100 37 6: 15! = 54,264 (AM) 1.1 b) Answer here. As a fraction, 0.0000184 is the probability that you will win the grand prize in the lottery in the next draw with these numbers 01 (b) The probability of winning the grand prize is :- As, the number of successful outcome is = 1 and we got from (a) that for six random sample Apolo pedimos the outcomes is = 54,264 ways. 01, 81, FL. 21 redmine est ar fed ts hv 00 1 =0.0000184. So, we can say that the wining probability is - 54,264 Fins
Student Number:21448762 STA1DCT Assignment 4 Name: Khalid Hasan Rohan c) Answer here. 0.999981 is the probability that you will not win the lottery in the next draw with these numbers 01 (c) The probability of not wining the grand prize with the chosen numbers is simply 1 minus the probability of winning :- According to ques: 1 - 1 94,263_or, 0.999981. 54,264 = 54,264 (Ans) d) Answer here. 3875 combinations are there available that include the numbers 20 and 21 but not the numbers 16, 17, 18 and 19. 7 01 (d) The numbers 20 and 21 are fixed in each combinations so we need to choose 4 more number Is from the remains 19, but not the number 16, 17,18 and 19. So, according to question the number of ways to choose from 4 numbers from 19: is : nCn = n=19 2XXXXX 00X10 n:(m-R!) n =4 = 19: 120,00 4. 151 = 19X18x17×16×18! 4115! : 31 using broke will pricing = 931024 (2) 10 $24 /3876. So, therefore we have to subtract the number of combinations that include all four numbers along 20,021, therefore the total number of combinations that include the number ~ 20 and 21 but not the number 16,17,18, 19 is, - PAS DE Siw od · P800000. C (19, 4) -1 = 3875. Using bristo 50, 3875 possible combination of 6 balls that include the 20, 21 but not the number 16,17,18,19. .1800000.10 2
Student Number:21448762 STA1DCT Assignment 4 Name: Khalid Hasan Rohan Question 2: a) Answer here. 1. 0.25 is the probability that the card dealt is a Cup. 4) Solution a) There are 10 cups card out of 90 card (u) to coron at folt from if So, the probability that the card dealt is a Cup 10 40 - = 0.25. (And): 2 stairs Vi derunt b) ) Answer here. x= 0, 1,