Louis rolls a biased dice 100 times. The probability that he will roll a 4 is (2)/(5) The probability that he will roll a 5 is 0.23 How many more times would you expect Louis to roll a 4 than a 5 ?

Question

Louis rolls a biased dice 100 times. The probability that he will roll a 4 is (2)/(5) The probability that he will roll a 5 is 0.23 How many more times would you expect Louis to roll a 4 than a 5 ?

Answer 1

This question is related to probability and expected value. Since the dice is biased, the probability of getting each number isn't the same. The probability of rolling a 4 is 2/5 and the probability of rolling a 5 is 0.23.We know that:1. Expected value(E) of an event = (number of trials) * (probability of each event).2. The number of trials in this case Louis rolling the dice is 100 times.3. So the expected number of times Louis rolls a '4' = (number of trials * probability of getting a '4') =100 * 2/5 = 40 times,4. Similarly, the expected value / number of times Louis rolls a '5’ = (number of trials * probability of getting a '5') = 100 * 0.23 = 23 times.5. Hence, The extra times you would expect Louis to roll a 4 than a 5 = expected number of times Louis rolls a '4' - expected number of times Louis rolls a '5’= 40 - 23 = 17 times

Question

Louis rolls a biased dice 100 times. The probability that he will roll a 4 is (2)/(5) The probability that he will roll a 5 is 0.23 How many more times would you expect Louis to roll a 4 than a 5 ?

Answer

Answer

Explanation