document.write( "Question 1120948: A medical survey was conducted in order to establish the proportion which was infected with cancer. The results indicated that 40% of the population were suffering from cancer.A sample of 6 people was later taken and examined for the disease. Find the probability that the following outcomes were observed:\r
\n" ); document.write( "\n" ); document.write( "(i) Only one person had cancer \r
\n" ); document.write( "\n" ); document.write( "(ii) At most two people had cancer \r
\n" ); document.write( "\n" ); document.write( "(iii) At least two people had cancer \r
\n" ); document.write( "\n" ); document.write( "(iv) Three or four people had cancer
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Algebra.Com's Answer #736655 by rothauserc(4718)\"\" \"About 
You can put this solution on YOUR website!
this problem can be viewed as a binomial experiment, since there are two possible outcomes for the patient and the probability (0.40) stays constant
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\n" ); document.write( "binomial probability formula is
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\n" ); document.write( "Probability (P) (k successes in n trials) = nCk * p^k * (1-p)^(n-k), where nCk = n! / (k! * (n-k)!)
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\n" ); document.write( "p = 0.40
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\n" ); document.write( "(i) P (1 success in 6 trials) = 6C1 * (0.40)^1 * (1-0.40)^(6-1) = 0.1866
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\n" ); document.write( "(ii) P (k < or = 2 in 6 trials) = 0.5443
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\n" ); document.write( "(iii) P (k > or = 2 in 6 trials) = 0.7667
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\n" ); document.write( "(iv) P (k = 3 or 4 in 6 trials) = P(k = 3 in 6 trials) + P(k = 4 in 6 trials) = 0.2764 + 0.1382 = 0.4146
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