Bernoulli distribution

The production of solar cells produces 2% of defective cells. Assume the cells are independent and that a lot contains 800 cells. Approximate the probability that less than 20 cells are defective. (Answer to the nearest three decimals).

Final Answer:

p =  0.814

Step-by-step explanation:

C0​(800)=(0800​)=0!(800−0)!800!​=11​=1 C1​(800)=(1800​)=1!(800−1)!800!​=1800​=800 C2​(800)=(2800​)=2!(800−2)!800!​=2⋅1800⋅799​=319600 C3​(800)=(3800​)=3!(800−3)!800!​=3⋅2⋅1800⋅799⋅798​=85013600 C4​(800)=(4800​)=4!(800−4)!800!​=4⋅3⋅2⋅1800⋅799⋅798⋅797​=16938959800 C5​(800)=(5800​)=5!(800−5)!800!​=5⋅4⋅3⋅2⋅1800⋅799⋅798⋅797⋅796​=2696682400160 C6​(800)=(6800​)=6!(800−6)!800!​=6⋅5⋅4⋅3⋅2⋅1800⋅799⋅798⋅797⋅796⋅795​=357310418021200 C7​(800)=(7800​)=7!(800−7)!800!​=7⋅6⋅5⋅4⋅3⋅2⋅1800⋅799⋅798⋅797⋅796⋅795⋅794​=40529210272690400 C8​(800)=(8800​)=8!(800−8)!800!​=8⋅7⋅6⋅5⋅4⋅3⋅2⋅1800⋅799⋅798⋅797⋅796⋅795⋅794⋅793​=4017457968280435900 C9​(800)=(9800​)=9!(800−9)!800!​≈3.535×1020=353536301208678359200 C10​(800)=(10800​)=10!(800−10)!800!​≈2.796×1022=27964721425606458212720 C11​(800)=(11800​)=11!(800−11)!800!​≈2.008×1024=2008375447839009271640800 C12​(800)=(12800​)=12!(800−12)!800!​≈1.320×1026=132050685695414859610382600 C13​(800)=(13800​)=13!(800−13)!800!​≈8.004×1027=8004303102152839182537037600 C14​(800)=(14800​)=14!(800−14)!800!​≈4.499×1029=449956181528163174046903470800 C15​(800)=(15800​)=15!(800−15)!800!​≈2.357×1031=23577703912075750320057741869920 C16​(800)=(16800​)=16!(800−16)!800!​≈1.156×1033=1156781098186216500077832960492950 C17​(800)=(17800​)=17!(800−17)!800!​≈5.334×1034=53348022410470219768295355354498400 C18​(800)=(18800​)=18!(800−18)!800!​≈2.320×1036=2320638974855454559920847957920680400 C19​(800)=(19800​)=19!(800−19)!800!​≈9.551×1037=95512614649313971887268584373366951200 q=2%=1002​=501​=0.02 n=800 d=20  p0​=(0n​)⋅ q0⋅ (1−q)n−0=1⋅ 0.020⋅ (1−0.02)800−0=9.5689⋅10−8 p1​=(1n​)⋅ q1⋅ (1−q)n−1=800⋅ 0.021⋅ (1−0.02)800−1≐1.5623⋅10−6 p2​=(2n​)⋅ q2⋅ (1−q)n−2=319600⋅ 0.022⋅ (1−0.02)800−2≐1.2737⋅10−5 p3​=(3n​)⋅ q3⋅ (1−q)n−3=85013600⋅ 0.023⋅ (1−0.02)800−3≐0.0001 p4​=(4n​)⋅ q4⋅ (1−q)n−4=16938959800⋅ 0.024⋅ (1−0.02)800−4≐0.0003 p5​=(5n​)⋅ q5⋅ (1−q)n−5=2696682400160⋅ 0.025⋅ (1−0.02)800−5≐0.0009 p6​=(6n​)⋅ q6⋅ (1−q)n−6=357310418021200⋅ 0.026⋅ (1−0.02)800−6≐0.0025 p7​=(7n​)⋅ q7⋅ (1−q)n−7=40529210272690400⋅ 0.027⋅ (1−0.02)800−7≐0.0057 p8​=(8n​)⋅ q8⋅ (1−q)n−8=4017457968280435900⋅ 0.028⋅ (1−0.02)800−8≐0.0116 p9​=(9n​)⋅ q9⋅ (1−q)n−9=353536301208678359200⋅ 0.029⋅ (1−0.02)800−9≐0.0208 p10​=(10n​)⋅ q10⋅ (1−q)n−10=27964721425606458212720⋅ 0.0210⋅ (1−0.02)800−10≐0.0335 p11​=(11n​)⋅ q11⋅ (1−q)n−11=2008375447839009271640800⋅ 0.0211⋅ (1−0.02)800−11≐0.0492 p12​=(12n​)⋅ q12⋅ (1−q)n−12=132050685695414859610382600⋅ 0.0212⋅ (1−0.02)800−12≐0.066 p13​=(13n​)⋅ q13⋅ (1−q)n−13=8004303102152839182537037600⋅ 0.0213⋅ (1−0.02)800−13≐0.0816 p14​=(14n​)⋅ q14⋅ (1−q)n−14=449956181528163174046903470800⋅ 0.0214⋅ (1−0.02)800−14≐0.0936 p15​=(15n​)⋅ q15⋅ (1−q)n−15=23577703912075750320057741869920⋅ 0.0215⋅ (1−0.02)800−15≐0.1001 p16​=(16n​)⋅ q16⋅ (1−q)n−16=1156781098186216500077832960492950⋅ 0.0216⋅ (1−0.02)800−16≐0.1002 p17​=(17n​)⋅ q17⋅ (1−q)n−17=53348022410470219768295355354498400⋅ 0.0217⋅ (1−0.02)800−17≐0.0943 p18​=(18n​)⋅ q18⋅ (1−q)n−18=2320638974855454559920847957920680400⋅ 0.0218⋅ (1−0.02)800−18≐0.0837 p19​=(19n​)⋅ q19⋅ (1−q)n−19=95512614649313971887268584373366951200⋅ 0.0219⋅ (1−0.02)800−19≐0.0703  p=p0​+p1​+p2​+p3​+p4​+p5​+p6​+p7​+p8​+p9​+p10​+p11​+p12​+p13​+p14​+p15​+p16​+p17​+p18​+p19​=9.5689⋅10−8+1.5623⋅10−6+1.2737⋅10−5+0.0001+0.0003+0.0009+0.0025+0.0057+0.0116+0.0208+0.0335+0.0492+0.066+0.0816+0.0936+0.1001+0.1002+0.0943+0.0837+0.0703=0.814

Try another example


Help us improve! If you spot a mistake, please let let us know. Thank you!







Tips for related online calculators
Looking for a statistical calculator?
Would you like to compute the count of combinations?

You need to know the following knowledge to solve this word math problem:

statisticscombinatoricsGrade of the word problem

Related math problems and questions: