A pair of dice is thrown once. Find the probability of getting a total of : (i) At least 9. (ii) Doublet of number (iii) Doublet of prime number.
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Answer:In a simultaneous throw of a pair of dice, probability of getting:(i) 8 as the sum =5/36(ii) a doublet =1/6(iii) a doublet of prime numbers =1/12(iv) a doublet of odd numbers =1/12(v) a sum greater than 9 =5/36(vi) an even number on first = 1/2(vii) an even number on one and a multiple of 3 on the other =11/36(viii) neither 9 nor 11 as the sum of the numbers on the faces =5/6(ix) a sum less than 6 =5/18(x) a sum less than 7 =7/18(xi) a sum more than 7 =11/18(xii) at least once = data insufficient(xiii) a number other than 5 in any dice.=25/36(xiv) even number on each die=1/4(xv) 5 as the sum=1/9(xvi) 2 will come up at least once=11/36(xvii) 2 will not come either time=25/36Step 1:When a pair of dice is thrown simultaneously total number of outcome is 6*6=36.Step 2:Problem (i) :To get 8 as a sum outcomes are(2,6),(6,2),(3,5),(5,3),(4,4)Total number of outcome is 5There the probability of getting 8 as sum is =5/36Problem (ii) :To get doublet the outcomes are(1,1),(2,2),(me is 6.Therefore probability of getting doublet is 6/36=1/6.Problem (iii) :To get the doublet of prime number the outcomme number is 3/36=1/12.Problem (iv) :To get doublet of odd number the outcomes are (1,1), (3,3),(5,5).Total number of outcomes is 3 .Therefore probability of getting doublet of odd numbers is 3/36=1/12.Problem (v) :To get the sum greater than 9 the outcomes are (4 ,6),(6,4),(5,6),(6,5), (6,6).Total number of outcomes is 5.Therefore probability of getting sum greater than 9 is =5/36.Pget the even number on first dice the outcomes are (2,1),(2,2),(2,3), (2,4),(2, 6 outcomesSimilar for considering 4and 6 at first dice we get 6 outcomes for each case.Now total number of outcome is 6*3=18.Therefore probability of getting even number on first dice is 18/36=1/2.Problem (vii)To get even number on one dice and multiple of 3 on other dice the outcomes are(2,3),(2,6),(3,2),(6,2),(4,3),(4,6),(3,4),(6,4),(6, 3),(6,6),(3,6)Total number of outcome is 11.Therefore probability of getting even number in one dice and multiple of odd number in other dice is 11/36.Problem (vii) :To get neither 9nor 11 as sum number of faces we have to consider other than the sum of 9 and 11 number on the faces.So the outcomes of sum 9 and 11 are (3,6),(6,3),(4,5),(5,4),(5,6),(6,5).Here total number of outcomes are 6.Therefore probability of getting neither 9 nor 11 as sum on dice is 1-6/36=5/6.Problem (ix)To get the sum less than 6 the outcomes are (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1).Total number of outcomes is 10.Therefore probability of getting sum less than 6 is 10/36=5/18.Problem (x) :To get the sum less than 7 the outcomes are (1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2)Total number of outcomes is 14.Therefore probability of getting sum less than 7 is =14/36=7/18.Problem (xi) :To get the sum more than 7 the outcomes are other (1,1),(1,2),(1,3),(1,4), (1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2)Total number of outcomes is 36-14=22Therefore probability of getting sum more than 7 is 22/36=11/18.Problem (xiii)To get the number other than 5 the outcomes are other than (5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(1,5),(2,5),(3,5),(4,5),(6,5)Total number of outcomes is 36-11=25.Therefore probability of getting number other than 5 is 25/36Problem (xiv)To get even number in each die the outcomes are (2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)Total number of outcome is 9.Therefore probability of getting even number in each dice is =9/36=1/4.Problem (xv)To get the sum as 5 the outcomes are (1,4),(2,3),(3,2),(4,1).Total number of outcomes is 4Therefore probability of getting sum as 5 = 4/36 =1/9Problem (xvi)2 will come up at least once here the outcomes are(1,2),(2,2), (2,3),(2,4),(2,5),(2,6),(3,2),(4,2),(5,2),(6,2).Total number of outcome is 11Therefore probability =11/36.Problem (xvii)In this problem we have to calculate the probability of 2 will not come either.In the previous problem we