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(D) Let $X_i$ be the outcome of the $i$-th throw. We want the experiment to end at the $5$-th throw,meaning the $4$-th and $5$-th throws must be $4$,a

Vedclass · Accepted Answer

$\frac{175}{6^5}$

Vedclass · Answer

(D) Let $X_i$ be the outcome of the $i$-th throw. We want the experiment to end at the $5$-th throw,meaning the $4$-th and $5$-th throws must be $4$,and the $3$-rd throw must not be $4$ (otherwise it would have ended at the $4$-th throw). The possible sequences of length $5$ ending in $44$ without ending earlier are: $1$. $4, X_2, X_3, 4, 4$ where $X_2 
eq 4$ and $X_3 
eq 4$. Probability $= \frac{1}{6} 	imes \frac{5}{6} 	imes \frac{5}{6} 	imes \frac{1}{6} 	imes \frac{1}{6} = \frac{25}{6^5}$. $2$. $X_1, 4, X_3, 4, 4$ where $X_1 
eq 4$ and $X_3 
eq 4$. Probability $= \frac{5}{6} 	imes \frac{1}{6} 	imes \frac{5}{6} 	imes \frac{1}{6} 	imes \frac{1}{6} = \frac{25}{6^5}$. $3$. $X_1, X_2, X_3, 4, 4$ where $X_3 
eq 4$ and the sequence does not contain $44$ in the first $4$ throws. Actually,the condition is that the sequence ends at the $5$-th throw. This means the sequence must be of the form $S_1, S_2, S_3, 4, 4$ where $S_3 
eq 4$ and no $44$ occurs in the first $4$ positions. The valid sequences are: $(4, X_2, X_3, 4, 4)$ where $X_2, X_3 
eq 4$: $\frac{1}{6} 	imes \frac{5}{6} 	imes \frac{5}{6} 	imes \frac{1}{6} 	imes \frac{1}{6} = \frac{25}{6^5}$. $(X_1, 4, X_3, 4, 4)$ where $X_1, X_3 
eq 4$: $\frac{5}{6} 	imes \frac{1}{6} 	imes \frac{5}{6} 	imes \frac{1}{6} 	imes \frac{1}{6} = \frac{25}{6^5}$. $(X_1, X_2, X_3, 4, 4)$ where $X_3 
eq 4$ and $(X_1, X_2) 
eq (4, 4)$: There are $5 	imes 5 	imes 5 = 125$ such sequences for the first $3$ positions,but we must exclude $(4, 4, X_3)$ which is $5$ sequences. So $125 - 5 = 120$. Wait,the total probability is $\frac{25+25+125}{6^5} = \frac{175}{6^5}$.

List-$I$	List-$II$
$(A) P(\overline{A} \cup B)$	$(I) \frac{2}{3}$
$(B) P(\frac{A}{\overline{B}})$	$(II) \frac{11}{15}$
$(C) P(A \cup B)$	$(III) \frac{3}{5}$

In a random experiment,a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to

Similar Questions

If $A$ and $B$ are two independent events such that $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{5}$,then which of the following is true?

If two unbiased six-faced dice are thrown simultaneously until a sum of either $7$ or $11$ occurs,then the probability that $7$ comes before $11$ is

$A$ and $B$ are two independent events. $P(A)=\frac{2}{5}, P(B)=\frac{1}{3}$. Match the following List-$I$ with List-$II$.
List-$I$ List-$II$
$(A) P(\overline{A} \cup B)$ $(I) \frac{2}{3}$
$(B) P(\frac{A}{\overline{B}})$ $(II) \frac{11}{15}$
$(C) P(A \cup B)$ $(III) \frac{3}{5}$

For independent events $A$ and $B$,$P(A \cup B) =$ . . . . . . .

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