A robot has a probability to solve a puzzle of 47% if the robot tries to solve 10 puzzles. 1. What is the probability that the robot to solve less than 2 puzzles ( not including 2)? Use 4 dig 2. On average, how many puzzles will be solved by the robot:

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hysgubwyri3 · Accepted Answer

Step 1  A random variable will follow the binomial distribution if it follows the conditions given below.  1. There are total 2 possible outcomes.  2. Total number of trials is fixed.  3. Each trial is independent.  4. Probability of success in each trial is equal.  The probability mass function for binomial distribution is given by P(x)=n!x!⋅(n−x)!⋅(p)x⋅(1−p)n−x, where x is number of successful outcomes, n is total number of trials and p is the probability of success. The expected value for this distribution is given by E(x)=n⋅p.  Step 2  The probability that the robot solves a puzzle is 0.47. The number of puzzles solved is a binomial random variable because it satisfies the condition as follows.  1. There are 2 possible outcomes, either the puzzle will be solved or not.  2. There are total 10 puzzles, so number of trials are fixed.  3. Each puzzle is independent to other.  4. Probability of solving each puzzle is 0.47.  So to write the probability mass function of number of puzzles solved, substitute n=10 and p=0.47 ∈ P(x)=n!x!⋅(n−x)!⋅(p)x⋅(1−p)n−x  P(x)=10!x!⋅(10−x)!⋅(0.47)x⋅(1−0.47)10−x  =10!x!⋅(10−x)!⋅(0.47)x⋅(0.53)10−x  Step 3-a  It is asked to find the probability of x&amp;lt;2 which will be equal to sum of probability of x=0 and x=1. To find the probability of x=0 and x=1, substitute these values in probability mass function.  P(x&amp;lt;2)=P(x=0)+P(x=1)  =10!0!⋅(10−0)!⋅(0.47)0⋅(0.53)10−0+10!1!⋅(10−1)!⋅(0.47)1⋅(0.53)10−1  =10!1⋅10!⋅1⋅(0.00175)+10⋅9!1⋅9!⋅(0.47)⋅(0.0033)  =0.00175+0.0155  ≈0.0173  So probability of solving less than 2 puzzles is approximately 0.0173.  Step 4-b  The expected or average value of binomial random variable is given by E(x)=n⋅p. So substitute n=10 and p=0.47 to find the average number of puzzles to be solved.  E(x)=10⋅0.47  =4.7  So on average 4.7 puzzles will be solved.

Stuart Rountree · Accepted Answer

Step 1&amp;nbsp;So, to write a mass function for the probability of the number of solved puzzles, substitute:&amp;nbsp;n=10&amp;nbsp;and&amp;nbsp;p=0.47&amp;nbsp;in:&amp;nbsp;P(x)=n!x!(n−x)!×(p)x×(1−p)n−x&amp;nbsp;Substituting values:&amp;nbsp;P(x)=10!x!(10−x)!×(0.47)x×(1−0.47)10−x&amp;nbsp;=10!x!(10−x)1×(0.47)x×(0.53)10−x&amp;nbsp;Your is to find the probability&amp;nbsp;x&amp;lt;2, which will be equal to the sum of the probabilities&amp;nbsp;x=0&amp;nbsp;and&amp;nbsp;x=1. To find the probability&amp;nbsp;x=0&amp;nbsp;and&amp;nbsp;x=1, substitute these values ​​into the probability mass function.&amp;nbsp;P(x&amp;lt;2)=P(x=0)+P(x=1)&amp;nbsp;=10!0!(10−0)!×(0.47)0×(0.53)10−0+10!1!(10−1)!×(0.47)1×(0.53)10−1&amp;nbsp;=10!1−10!×1×(0.00175)+10−9!1−9!×(0.47)×(0.0033)&amp;nbsp;=0.00175+0.0155&amp;nbsp;≈0.0173&amp;nbsp;Thus, the probability of solving fewer than two puzzles is approximately 0.0173.&amp;nbsp;Answer b:&amp;nbsp;E(x)=10×0.47&amp;nbsp;=4.7&amp;nbsp;So on average 4.7 puzzles will be solved.

A robot has a probability to solve a puzzle of 47% if the ro

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