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

Probability

Question
CBSEENMA12033870
Wired Faculty App

In four throws with a pair of dice, what is the probability of throwing doublets at least twice?

Solution

We get a doublet when we get
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Let p be probability of getting a doublet
therefore space straight p space equals space 6 over 36 space equals space 1 over 6 space space space space space space straight q space equals space 1 minus straight p space equals space 1 space minus 1 over 6 space equals space 5 over 6
P(at least two doublets) = 1 - [P(0) + P(1)]     
                  equals space 1 minus open square brackets straight C presuperscript 4 subscript 0 open parentheses 1 over 6 close parentheses to the power of 0 space open parentheses 5 over 6 close parentheses to the power of 4 space plus space straight C presuperscript 4 subscript 1 space open parentheses 1 over 6 close parentheses to the power of 1 space open parentheses 5 over 6 close parentheses cubed close square brackets space space space space space space space space open square brackets because space space straight P left parenthesis straight r right parenthesis space equals space straight C presuperscript straight n subscript straight r space straight p to the power of straight r space straight q to the power of straight n minus straight r end exponent comma space straight n space equals space 4 close square brackets equals space 1 minus open square brackets 1 cross times 1 cross times 625 over 1296 plus 4 over 1 cross times 1 over 6 cross times 125 over 216 close square brackets space equals space 1 minus open square brackets 625 over 1296 plus 500 over 1296 close square brackets space equals space 1125 over 1296 equals space space fraction numerator 1296 minus 1125 over denominator 1296 end fraction space equals space 171 over 1296

Some More Questions From Probability Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation