Question

Suppose a simple random sample of 90 students from this district is drawn. What is the probability that at least half of them have either brown or green eyes? (Use technology. Round your answer to three decimal places.)

brown eyes is 40%, green eyes is 20%.

Answer 1

Sample size , n =    90
Probability of an event of interest, p = brown + green = 0.4 + 0.2

=0.6

	X	P(X)
P ( X = 45) = C (90,45) * 0.6^45 * ( 1 - 0.6)^45=	45	0.0134
P ( X = 46) = C (90,46) * 0.6^46 * ( 1 - 0.6)^44=	46	0.0196
P ( X = 47) = C (90,47) * 0.6^47 * ( 1 - 0.6)^43=	47	0.0275
P ( X = 48) = C (90,48) * 0.6^48 * ( 1 - 0.6)^42=	48	0.0370
P ( X = 49) = C (90,49) * 0.6^49 * ( 1 - 0.6)^41=	49	0.0476
P ( X = 50) = C (90,50) * 0.6^50 * ( 1 - 0.6)^40=	50	0.0585
P ( X = 51) = C (90,51) * 0.6^51 * ( 1 - 0.6)^39=	51	0.0688
P ( X = 52) = C (90,52) * 0.6^52 * ( 1 - 0.6)^38=	52	0.0774
P ( X = 53) = C (90,53) * 0.6^53 * ( 1 - 0.6)^37=	53	0.0833
P ( X = 54) = C (90,54) * 0.6^54 * ( 1 - 0.6)^36=	54	0.0856
P ( X = 55) = C (90,55) * 0.6^55 * ( 1 - 0.6)^35=	55	0.0840
P ( X = 56) = C (90,56) * 0.6^56 * ( 1 - 0.6)^34=	56	0.0788
P ( X = 57) = C (90,57) * 0.6^57 * ( 1 - 0.6)^33=	57	0.0705
P ( X = 58) = C (90,58) * 0.6^58 * ( 1 - 0.6)^32=	58	0.0602
P ( X = 59) = C (90,59) * 0.6^59 * ( 1 - 0.6)^31=	59	0.0489
P ( X = 60) = C (90,60) * 0.6^60 * ( 1 - 0.6)^30=	60	0.0379
P ( X = 61) = C (90,61) * 0.6^61 * ( 1 - 0.6)^29=	61	0.0280
P ( X = 62) = C (90,62) * 0.6^62 * ( 1 - 0.6)^28=	62	0.0196
P ( X = 63) = C (90,63) * 0.6^63 * ( 1 - 0.6)^27=	63	0.0131
P ( X = 64) = C (90,64) * 0.6^64 * ( 1 - 0.6)^26=	64	0.0083
P ( X = 65) = C (90,65) * 0.6^65 * ( 1 - 0.6)^25=	65	0.0050
P ( X = 66) = C (90,66) * 0.6^66 * ( 1 - 0.6)^24=	66	0.0028
P ( X = 67) = C (90,67) * 0.6^67 * ( 1 - 0.6)^23=	67	0.0015
P ( X = 68) = C (90,68) * 0.6^68 * ( 1 - 0.6)^22=	68	0.0008
P ( X = 69) = C (90,69) * 0.6^69 * ( 1 - 0.6)^21=	69	0.0004
P ( X = 70) = C (90,70) * 0.6^70 * ( 1 - 0.6)^20=	70	0.0002
P ( X = 71) = C (90,71) * 0.6^71 * ( 1 - 0.6)^19=	71	0.0001
P ( X = 72) = C (90,72) * 0.6^72 * ( 1 - 0.6)^18=	72	0.0000
P ( X = 73) = C (90,73) * 0.6^73 * ( 1 - 0.6)^17=	73	0.0000
P ( X = 74) = C (90,74) * 0.6^74 * ( 1 - 0.6)^16=	74	0.0000
P ( X = 75) = C (90,75) * 0.6^75 * ( 1 - 0.6)^15=	75	0.0000
P ( X = 76) = C (90,76) * 0.6^76 * ( 1 - 0.6)^14=	76	0.0000
P ( X = 77) = C (90,77) * 0.6^77 * ( 1 - 0.6)^13=	77	0.0000
P ( X = 78) = C (90,78) * 0.6^78 * ( 1 - 0.6)^12=	78	0.0000
P ( X = 79) = C (90,79) * 0.6^79 * ( 1 - 0.6)^11=	79	0.0000
P ( X = 80) = C (90,80) * 0.6^80 * ( 1 - 0.6)^10=	80	0.0000
P ( X = 81) = C (90,81) * 0.6^81 * ( 1 - 0.6)^9=	81	0.0000
P ( X = 82) = C (90,82) * 0.6^82 * ( 1 - 0.6)^8=	82	0.0000
P ( X = 83) = C (90,83) * 0.6^83 * ( 1 - 0.6)^7=	83	0.0000
P ( X = 84) = C (90,84) * 0.6^84 * ( 1 - 0.6)^6=	84	0.0000
P ( X = 85) = C (90,85) * 0.6^85 * ( 1 - 0.6)^5=	85	0.0000
P ( X = 86) = C (90,86) * 0.6^86 * ( 1 - 0.6)^4=	86	0.0000
P ( X = 87) = C (90,87) * 0.6^87 * ( 1 - 0.6)^3=	87	0.0000
P ( X = 88) = C (90,88) * 0.6^88 * ( 1 - 0.6)^2=	88	0.0000
P ( X = 89) = C (90,89) * 0.6^89 * ( 1 - 0.6)^1=	89	0.0000
P ( X = 90) = C (90,90) * 0.6^90 * ( 1 - 0.6)^0=	90	0.0000

Probability = 0.979

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