Multiple choice

Find the minimum number of tosses of a pair of dice so that the probability of getting the sum of the digits on the dice equal to 7 on at least one toss is greater than $0.95$($\log _{10 }{2 } =0.3010;\log _{10 }{3=0.4771 } $)

  1. $17$
  2. $18$
  3. $19$
  4. $20$
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A Correct answer
Explanation

Probability of sum 7 with two dice = 6/36 = 1/6. Probability of not getting 7 = 5/6. Probability of not getting 7 in n tosses = (5/6)^n. We need 1 - (5/6)^n > 0.95, so (5/6)^n < 0.05. Taking logs: n * (log 5 - log 6) < log 0.05. n * (0.699 - 0.778) < -1.301. n * (-0.079) < -1.301. n > 16.46. Minimum n = 17.

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