Begin2.DVI

larger than 1 because if 2 items or more are defective out of 6,we cannot obtain 5

nondefectives.P(X≤1) = 1/11 + 9/22 = 1/ 2

Ifn= 7items are selectedf(x) =

( 3

x

)( 9

7 −x

)

( 12

7

). Here

f(0) = 1/ 22 , f(1) = 7/ 22 , f(2) = 21/ 44

thenP(X≤2) =

1

22 +

7

22 +

21

44 =

37

44 = 0.^84 >^0.^8

I11-24. (b)(p+q)^5 =p^5 + 5p^4 q+ 10p^3 q^2 + 10p^2 q^3 + 5pq^4 +q^5

(c)

1

32

=p^5 = (^1

2

)^5 probability of 5 heads

5

32 =5p

(^4) q= 5(^1
2 )
(^4) (^1
2 )probability of 4 heads, 1 tail
10
32
=10p^3 q^2 = 10(^1
2
)^3 (^1
2
)^2 probability of 3 heads, 2 tails
10
32
=10p^2 q^3 = 10(^1
2
)^2 (^1
2
)^3 porobability of 2 heads, 1 tail
13
16 =p
5
5 p
(^4) q+ 10p (^3) q (^2) + 10p (^2) q (^3) probability of getting at least 2 heads
I11-25.
x f(x) =
( 40
x
)
(0.8)x(0.2)^40 −x
40 0.000132923
39 0.00132923
38 0.00647999
37 0.02052
36 0.0474524
35 0.0854143
34 0.124563
33 0.151255
32 0.155981
31 0.13865
(a)f(33) =P(X= 33) = 0. 151255
(b)P(X= 37) =f(37) = 0. 02052
(c)
∑ 37
k=0= 1−f(40)−f(39)−f(38) = 0.^992057963
(d)
∑ 40
k=32f(k) =f(32) +···+f(40) = 0.^59312713
Solutions Chapter 11