Data analysis

Given a fair 7-sided die with numbers 1 to 7.

It is rolled 5 times.

So total number of outcomes \(= 7^{5}\)

as one time rolled can give any number from 1 to \(7\ \Rightarrow\ 7\) out comes

So five time rolled \(= 7^{5} = 16.807\)

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 \(\Rightarrow\ 5\) ways

And in remaining 4 rolls, any outcome can come \(\Rightarrow 7^{4}\) ways

Total favourable ways \(= 5\ \times\ 7^{4}\)

Probability \(= (5\ \times\ 7^{4})/7^{5}\)

\(= 5/7\)

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 \(\Rightarrow\) possible outcomes are 1, 2, 4, 5, 6, 7

Number of ways \(= 6\)

For five time rolls \(= 5^{6}\)

Probability \(= 1\ -\ (5^{6}/5^{7})\)

\(= 1\ -\ (1/5)\)

\(= 4/5\)

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 \(\Rightarrow\ 2,\ 4,\ 6,\ \Rightarrow\) 3 outcomes

So 4 rolls even \(= 4^{3}\) ways

And remaining one roll in 7 ways

Total favourable ways \(= 7\ \times\ 4^{3}\)

Probability \(= (7\ \times\ 4^{3})/7^{5}\)

\(= 64/2401

Hence the probability that exactly 4 of the rolls show even number.

(Rounded to 3 decimals)

Given a fair 7-sided die with numbers 1 to 7.

It is rolled 5 times.

So total number of outcomes \(= 7^{5}\)

as one time rolled can give any number from 1 to \(7\ \Rightarrow\ 7\) out comes

So five time rolled \(= 7^{5} = 16.807\)

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 \(\Rightarrow\ 5\) ways

And in remaining 4 rolls, any outcome can come \(\Rightarrow 7^{4}\) ways

Total favourable ways \(= 5\ \times\ 7^{4}\)

Probability \(= (5\ \times\ 7^{4})/7^{5}\)

\(= 5/7\)

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 \(\Rightarrow\) possible outcomes are 1, 2, 4, 5, 6, 7

Number of ways \(= 6\)

For five time rolls \(= 5^{6}\)

Probability \(= 1\ -\ (5^{6}/5^{7})\)

\(= 1\ -\ (1/5)\)

\(= 4/5\)

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 \(\Rightarrow\ 2,\ 4,\ 6,\ \Rightarrow\) 3 outcomes

So 4 rolls even \(= 4^{3}\) ways

And remaining one roll in 7 ways

Total favourable ways \(= 7\ \times\ 4^{3}\)

Probability \(= (7\ \times\ 4^{3})/7^{5}\)

\(= 64/2401

Hence the probability that exactly 4 of the rolls show even number.

(Rounded to 3 decimals)