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Question
CBSEENMA12033913
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Suppose X has a binomial distribution Bopen parentheses 6 comma space 1 half close parentheses. Show that X = 3 is the most likely outcome. 

Solution

Since straight B space open parentheses 6 comma space 1 half close parentheses is binomial distribution
        therefore space space space space space straight n space equals space 6 comma space space space straight p space equals space 1 half comma space space straight q space equals space 1 space minus space straight p space equals space 1 minus space 1 half space equals space 1 half
                       straight P left parenthesis straight X space equals space 0 right parenthesis space equals space straight C presuperscript 6 subscript 0 space open parentheses 1 half close parentheses to the power of 6 space equals space 1 space cross times space 1 over 64 space equals 1 over 64 straight P left parenthesis straight X space equals space 1 right parenthesis space equals space straight C presuperscript 6 subscript 1 space open parentheses 1 half close parentheses to the power of 5 space open parentheses 1 half close parentheses space equals space 6 space cross times space 1 over 32 cross times space 1 half space equals space 6 over 64 straight P left parenthesis straight X space equals space 2 right parenthesis space equals space straight C presuperscript 6 subscript 2 space open parentheses 1 half close parentheses to the power of 4 space open parentheses 1 half close parentheses squared space equals fraction numerator 6 space cross times space 5 over denominator 1 space cross times 2 end fraction cross times 1 over 16 cross times 1 fourth space equals 15 over 64 straight P left parenthesis straight X space equals 3 right parenthesis space equals space straight C presuperscript 6 subscript 3 space open parentheses 1 half close parentheses cubed space open parentheses 1 half close parentheses cubed space equals space fraction numerator 6 space cross times space 5 space cross times space 4 over denominator 1 space cross times space 2 space cross times 3 end fraction cross times space 1 over 8 cross times 1 over 8 space equals space 20 over 64 straight P left parenthesis straight X space equals space 4 right parenthesis space equals space straight C presuperscript 6 subscript 4 space open parentheses 1 half close parentheses squared space open parentheses 1 half close parentheses to the power of 4 space equals fraction numerator 6 space cross times 5 over denominator 1 cross times 2 end fraction cross times 1 fourth cross times 1 over 16 equals 15 over 64 straight P left parenthesis straight X space equals 5 right parenthesis space equals space straight C presuperscript 6 subscript 5 space open parentheses 1 half close parentheses space open parentheses 1 half close parentheses to the power of 5 space equals space 6 over 1 cross times 1 half cross times 1 over 32 equals space 6 over 64 straight P left parenthesis straight X space equals 6 right parenthesis space equals space straight C presuperscript 6 subscript 6 space open parentheses 1 half close parentheses to the power of 6 space equals space 1 space cross times space 1 over 64 space equals 1 over 64.
Since 20 over 64 is maximum of all the above values
therefore space space space straight X space equals space 3 space is space the space most space likely space out space come. space

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