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If I throw 3 standard 7-sided dice, what is the probability that the sum of their top faces equals to 21? Assume both throws are independent to each other.(a) \(\frac{1}{273}\)(b) \(\frac{2}{235}\)(c) \(\frac{1}{65}\)(d) \(\frac{2}{9}\)I have been asked this question in homework.I'm obligated to ask this question of Multiplication Theorem on Probability in chapter Discrete Probability of Discrete Mathematics

Answer»

The CORRECT CHOICE is (a) \(\frac{1}{273}\)

The best I can explain: To obtain a sum of 21 from three 7-sided dice is that 3 die will SHOW 7 face up. THEREFORE, the probability is simply \(\frac{1}{7} * \frac{1}{7} * \frac{1}{7} = \frac{1}{273}\).



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