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The probability of a successful optical alignment in the assembly of an optical data storage product is 0.8. Assume the trials are independent.

a. What is the probability that the first successful alignment requires exactly four trials?

b. What is the probability that the first successful alignment requires at most four trials?

c. What is the probability that the first successful alignment requires at least four trials?

yuhki  Nov 21, 2014

Best Answer 

 #1
avatar+91465 
+5

a) P(FFFS)=0.2^3*0.8 

$${{\mathtt{0.2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{0.8}} = {\frac{{\mathtt{4}}}{{\mathtt{625}}}} = {\mathtt{0.006\: \!4}}$$

 

b) 1-P(FFFF) = 1-0.2^4 

$${\mathtt{1}}{\mathtt{\,-\,}}{{\mathtt{0.2}}}^{{\mathtt{4}}} = {\frac{{\mathtt{624}}}{{\mathtt{625}}}} = {\mathtt{0.998\: \!4}}$$

 

c) P(FFF)=0.2^3

$${{\mathtt{0.2}}}^{{\mathtt{3}}} = {\frac{{\mathtt{1}}}{{\mathtt{125}}}} = {\mathtt{0.008}}$$

 

That is what I think anyway.  :))

Perhaps someone would like to check my answers?

Melody  Nov 22, 2014
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1+0 Answers

 #1
avatar+91465 
+5
Best Answer

a) P(FFFS)=0.2^3*0.8 

$${{\mathtt{0.2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{0.8}} = {\frac{{\mathtt{4}}}{{\mathtt{625}}}} = {\mathtt{0.006\: \!4}}$$

 

b) 1-P(FFFF) = 1-0.2^4 

$${\mathtt{1}}{\mathtt{\,-\,}}{{\mathtt{0.2}}}^{{\mathtt{4}}} = {\frac{{\mathtt{624}}}{{\mathtt{625}}}} = {\mathtt{0.998\: \!4}}$$

 

c) P(FFF)=0.2^3

$${{\mathtt{0.2}}}^{{\mathtt{3}}} = {\frac{{\mathtt{1}}}{{\mathtt{125}}}} = {\mathtt{0.008}}$$

 

That is what I think anyway.  :))

Perhaps someone would like to check my answers?

Melody  Nov 22, 2014

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