Marks will be awarded for accuracy in the rounding of final answers where you are asked to round. To ensure that you receive these marks, take care in

Globokim8

Globokim8

Answered question

2020-12-29

Marks will be awarded for accuracy in the rounding of final answers where you are asked to round. To ensure that you receive these marks, take care in keeping more decimals in your intermediate steps than what the question is asking you to round your final answer to.
A fair 7 -sided die with the numbers 1 trough 7 is rolled five times. Express each of your answers as a decimal rounded to 3 decimal places.
(a) What is the probability that exactly one 3 is rolled?
(b) What is the probability that at least one 3 is rolled?
(c) What is the probability that exactly four of the rolls show an even number?

Answer & Explanation

Talisha

Talisha

Skilled2020-12-30Added 93 answers

Data analysis
Given a fair 7-sided die with numbers 1 to 7.
It is rolled 5 times.
So total number of outcomes =75
as one time rolled can give any number from 1 to 7  7 out comes
So five time rolled =75=16.807
To find the following probabilities.
Sub part a)
Probability that exactly one 3 is rolled.
One 3 can come in any of the 5 rolls  5 ways
And in remaining 4 rolls, any outcome can come 74 ways
Total favourable ways =5 × 74
Probability =(5 × 74)/75
=5/7
Hence the probability that only one 3 is rolled is 0.714
(Rounded to 3 decimals)
Sub part b)
Probability that at least one 3 is rolled
=1  (probability of 3 is never rolled)
So 3 should never come possible outcomes are 1, 2, 4, 5, 6, 7
Number of ways =6
For five time rolls =56
Probability =1  (56/57)
=1  (1/5)
=4/5
Hence the probability that atleast one three is rolled is 0.8
Sub part c)
Probability that exactly 4 of the rolls show even number.
Even number  2, 4, 6,  3 outcomes
So 4 rolls even =43 ways
And remaining one roll in 7 ways
Total favourable ways =7 × 43
Probability =(7 × 43)/75
\(= 64/2401
Hence the probability that exactly 4 of the rolls show even number.
(Rounded to 3 decimals)

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