Question

A game of telephone is played with 5 students: Angela, Bob, Carlos, Dee, and Frank. That is, a teacher whispers a saying to Angela, who whispers it Bob, who whispers is to Carlos, who whispers it to Dee, who whispers it to Frank, and Frank announces what he thinks he heard. Angela understands the correct saying 80% of the time, Bob’s accuracy is 65%, Carlos 85%, Dee 70%, and Frank 75%. Given that the saying at the end does not match the saying from the teacher, what is the probability that Bob was inaccurate? g

Answer 1

0.4558 is the required probability here.

Explanation:

We first compute the probability of the saying not reaching correctly here as:

= 1 - Probability that all of them said it correctly

= 1 - 0.8*0.65*0.85*0.7*0.75

= 0.76795

Now the probability that given the end saying does not match the one from the teacher, conditional probability that Bob was inaccurate is computed using Bayes theorem here as:

= P( Bob was inaccurate and saying did not reach correctly ) / P( saying did not reach correctly)

= (1 - 0.65) / 0.76795

= 0.4558

Therefore 0.4558 is the required probability here.

Answer 2

0.3646 or 36.46%

Explanation:

The probability that Bob was the first to be inaccurate is:

`P(B) = 0.8*(1-0.65)=0.28`

The probability that the statement ends up being inaccurate is 100% minus the probability of the saying being accurate at the end:

`P(I) = 1-P(A)\\P(I)=1-(0.8*0.65*0.85*0.7*0.75)\\P(I)=0.76795}`

The probability that Bob was inaccurate given that the saying does not match the saying from the teacher is:

`P(B|I) = (0.28)/(0.76795)\\P(B|I)=0.3646`

The probability is 0.3646 or 36.46%.