+91-8076753736[email protected]
logo
logo
-->

Probability

Question
CBSEENMA12033897
Wired Faculty App

A pair of dice is thrown 6 times. If getting a total of 8 is considered a success, what is the probability of 
(i) at least 4 successes?
(ii) at most 5 succcesses?

Solution

Here n = 6
 p = P{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) = 5 over 36
q = 1 - p = 1 - 5 over 36 equals space 31 over 36
(i) P(at least 4 successes) = P(4) + P(5)  + P(6)
                    equals space straight C presuperscript 6 subscript 4 space open parentheses 31 over 36 close parentheses squared space open parentheses 5 over 36 close parentheses to the power of 4 space plus space straight C presuperscript 6 subscript 5 space open parentheses 31 over 36 close parentheses space open parentheses 5 over 36 close parentheses to the power of 5 space plus space straight C presuperscript 6 subscript 6 space open parentheses 5 over 36 close parentheses to the power of 6 equals space open parentheses 5 over 36 close parentheses to the power of 4 space open square brackets fraction numerator 6 space cross times space 5 over denominator 1 space cross times 2 end fraction space cross times space fraction numerator 31 space cross times space 31 over denominator 36 space cross times space 36 end fraction plus space 6 over 1 cross times space 31 over 36 cross times space 5 over 36 plus 1 space cross times space 5 over 36 space cross times space 5 over 36 close square brackets equals space open parentheses 5 over 36 close parentheses to the power of 4 space open square brackets fraction numerator 15 space cross times space 31 space cross times space 31 over denominator 36 space cross times space 36 end fraction plus space fraction numerator 6 space cross times space 31 space cross times space 5 over denominator 36 space cross times space 36 end fraction space plus space fraction numerator 25 over denominator 36 space cross times space 36 end fraction close square brackets equals space fraction numerator left parenthesis 5 right parenthesis to the power of 4 over denominator left parenthesis 36 right parenthesis to the power of 4 end fraction left square bracket 15 space cross times space 31 space cross times space 31 space plus space 6 space cross times space 31 space cross times space 5 space plus space 25 right square bracket space equals space fraction numerator left parenthesis 5 right parenthesis to the power of 4 over denominator left parenthesis 36 right parenthesis to the power of 6 end fraction left square bracket 14415 plus 930 plus 25 right square bracket space equals fraction numerator left parenthesis 15370 right parenthesis thin space left parenthesis 5 right parenthesis to the power of 4 over denominator left parenthesis 36 right parenthesis to the power of 6 end fraction
(ii) P(at atmost 5 successess) = 1 - P(6) = 1 - straight C presuperscript 6 subscript 6open parentheses 5 over 36 close parentheses to the power of 6 = space 1 minus open parentheses 5 over 36 close parentheses to the power of 6

Some More Questions From Probability Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation