Question

9.55. We distribute 15 books among four students so that all allocations are equally likely. What is the probability that
1. the first student gets at least 2 books?
2. the first or the last student gets at least two books?
3. no one gets less than three books?

Answer 1

(9.55) solution : Given data, * We distribute 15 books among four students. So * That all allocat'ons are equally likely Let es consider students are * The number of books received by 2173,220, gets at least a books :- The first student * P(4,22 ): 1-P(2,-0) -p(x, =1). . Total case when 2,20 All possible cases

integer sted solution ie to solving non-negative linear equation, * all cases solution to itu 215 = x+&+dz taua since, ntr- 9, tytiriot xx ani! * 15+3. je 15 +4-1 num 4-1 gun *P(x, co ) 7 e soln. (ay +2 +24=15) soln (x+4+3, +4515) (5+3 - 1 15+41 .. 4-1

1572 15+2ica 15+3c 1762 cu 2551 136 I 846 6 /PC, 20 ) = -6 1. Solm of (a +2 +4 = 14 ) * P (2,57) - Solin of (a,+*+*+%4y= 15) 14+31 15+41

14 +26 15+3 1602 B03 Todos 816 34 * P (,22.) = 16 - 1-0.16667 -0.14705, 50. 68628:

(PCajza) -0.6863) (2) Alow we have *P(2,22 ) = 1-p(7430) -P(1431): similar proceed to case (1) and Now we can get the same result, There fore, PC 24.22) = 0.6863. *P(2,22 and 2422 ) = 1-p(x,-0 and 24-0) – p(,50 and 2 =0) -P(2,2 l aud 7, 50 ) - P(2, 31 and 24 =0)

al som to (a +8=15) Total cases Sohn to (x +ą = 14 ) . All cases soln to (22 +22=14 ) soln to (0+2= 13) All cases All cases ther 150, 15c, e 15 T 15 816 14 816 816 3 816-16-15-15-14 816 756 63

20.92647 : {p(7,22 and 94, 22 ) 20-926 * Plz, za or MP 22 ) = P(4,22 ) +p(2422] -PC, 22 and 94 22). = 0.6863 +0.6863 -0.926. a 0.4466. Hence, | P(3 23 BY 3 22 ) - 6.66.

(3) Now, 9,1g, 23, 24 23. 3 lie 2,% + 8 7x7 +12=15, a ta ta ta = 15-12 9, +4 + xy +24 213, where a, ul, %₂,24 7 0. solito (1,42,45+233) *P(2,23,923,523,4,23). Se . All cases 3th-tcut. - 1803

= .01024509 20.02451 Here PC2,23,323,323,223) -0.0245. - o-