When two dice are rolled, let x be the probab | vedclass

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(A) The total number of outcomes when two dice are rolled is $6 	imes 6 = 36$.
The outcomes where the sum is at most $7$ are:
Sum $= 2: (1,1)$ ($1$

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The total number of outcomes when two dice are rolled is $6 imes 6 = 36$. The outcomes where the sum is at most $7$ are: Sum $= 2: (1,1)$ ($1$ outcome) Sum $= 3: (1,2), (2,1)$ ($2$ outcomes) Sum $= 4: (1,3), (2,2), (3,1)$ ($3$ outcomes) Sum $= 5: (1,4), (2,3), (3,2), (4,1)$ ($4$ outcomes) Sum $= 6: (1,5), (2,4), (3,3), (4,2), (5,1)$ ($5$ outcomes) Sum $= 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ ($6$ outcomes) Total favorable outcomes $= 1+2+3+4+5+6 = 21$. Thus, $x = \frac{21}{36} = \frac{7}{12}$. For $y$, the probability of getting a sum of $7$ in one roll is $p = \frac{6}{36} = \frac{1}{6}$. The probability of not getting a sum of $7$ is $q = 1 - \frac{1}{6} = \frac{5}{6}$. When rolled $n$ times, the probability of getting a sum of $7$ at least once is $y = 1 - q^n = 1 - (\frac{5}{6})^n$. We want $y > x$, so $1 - (\frac{5}{6})^n > \frac{7}{12} \Rightarrow (\frac{5}{6})^n < 1 - \frac{7}{12} = \frac{5}{12}$. For $n=1: \frac{5}{6} \approx 0.833 > 0.416$ For $n=2: \frac{25}{36} \approx 0.694 > 0.416$ For $n=3: \frac{125}{216} \approx 0.578 > 0.416$ For $n=4: \frac{625}{1296} \approx 0.482 > 0.416$ For $n=5: \frac{3125}{7776} \approx 0.401 < 0.416$. Thus, the minimum $n$ is $5$.

When two dice are rolled, let $x$ be the probability that the sum of the numbers appearing on the dice is at most $7$. Let $y$ be the probability of getting a sum of $7$ at least once when a pair of dice are rolled $n$ times. In order to have $y > x$, the minimum value of $n$ is

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