document.write( "Question 1174089: Question no 2 (A): Students are randomly selected from SMIU, probability that they are from BA is ½, probability that they are from CS is 1/3 and remaining is the probability that they are from Media Studies department. Group of 12 students is selected:
\n" ); document.write( "(a) What is the probability that at least 6 are from BA department
\n" ); document.write( "(b) What is the probability that none is from media department
\n" ); document.write( "(c) What is the probability that at most three are from CS Department
\n" ); document.write( "Question No 2 (B): Assume that the number of road accidents in any month are no ofcharacters in your first name. What is the probability fewer than three accidents will occur in a particular month?
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Algebra.Com's Answer #799486 by Boreal(15235) $\"\"$ $\"About$

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2B is a Poisson distribution with lambda= to the number of characters in your first name. P(x< =3) would be poissoncdf(char.,2)ENTER. The \"2\" is the largest number of accidents below 3 and includes 0 and 1 but not 3.
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\n" ); document.write( "p(0-5) is 0.3872, so 1-that probability is the answer or 0.6128
\n" ); document.write( "p(Media Dept ) is 1/6 so probability none is from median dept is (5/6)^12=0.1122
\n" ); document.write( "this is p (0,1,2,3):p(0) is (2/3)^12=0.0078; p(1) is 12*(2/3)^11*(1/3)=0.0424; 66*(2/3)^10*(1/9)=0.1272, the 66 coming from 12C2; and p(3)=220*(2/3)^9*(1/27)=0.2120
\n" ); document.write( "total is 0.3931. Also from binomialcdf(12, (1/3),3) ENTER. \n" ); document.write( "

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