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(D) When two dice are thrown,the total number of outcomes is $6 	imes 6 = 36$. We need to find the number of pairs $(x, y)$ such that $x^2 + y^2 \leq

Vedclass · Accepted Answer

$\frac{5}{12}$

Vedclass · Answer

(D) When two dice are thrown,the total number of outcomes is $6 	imes 6 = 36$. We need to find the number of pairs $(x, y)$ such that $x^2 + y^2 \leq 25$,where $x, y \in \{1, 2, 3, 4, 5, 6\}$. Let us list the favorable outcomes for each value of $x$: If $x = 1$,then $1^2 + y^2 \leq 25 \implies y^2 \leq 24$. Possible $y$ values are $\{1, 2, 3, 4\}$ ($4$ outcomes). If $x = 2$,then $2^2 + y^2 \leq 25 \implies y^2 \leq 21$. Possible $y$ values are $\{1, 2, 3, 4\}$ ($4$ outcomes). If $x = 3$,then $3^2 + y^2 \leq 25 \implies y^2 \leq 16$. Possible $y$ values are $\{1, 2, 3, 4\}$ ($4$ outcomes). If $x = 4$,then $4^2 + y^2 \leq 25 \implies y^2 \leq 9$. Possible $y$ values are $\{1, 2, 3\}$ ($3$ outcomes). If $x = 5$,then $5^2 + y^2 \leq 25 \implies y^2 \leq 0$. No possible $y$ values since $y \geq 1$. If $x = 6$,then $6^2 + y^2 \leq 25$. No possible $y$ values. Total favorable outcomes $= 4 + 4 + 4 + 3 = 15$. The probability is $\frac{15}{36} = \frac{5}{12}$. Thus,option $(d)$ is correct.

List $I$	List $II$
$(i) \ P(A \cap B)$	$(1) \ 0.4$
$(ii) \ P(A \cap \bar{B})$	$(2) \ 0.2$
$(iii) \ P(\bar{A} \cap B)$	$(3) \ 0.3$
$(iv) \ P(\bar{A} \cap \bar{B})$	$(4) \ 0.1$

When two dice are thrown,the probability of getting an ordered pair $(x, y)$ such that $x^2+y^2 \leq 25$,where $x$ and $y$ are the numbers that show up on the two dice,is:

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