Multiplication Rule in Probability

[odolwa99]

Homework Statement

In order to highlight the problem I'm having here I have posted two questions, the answer for the second matches the textbook answer, the first does not. The wording of both questions appears to be the same. Have I gone wrong in the 1st question, or is the book incorrect?

Many thanks.

Q. 1. Three people were selected at random and asked on which day of the week their next birthday was falling. What is the probability that only one of the birthdays falls on a Sunday.

Q. 2. A fair die is thrown 3 times. Find the probability that there will be exactly one 6.

Homework Equations

The Attempt at a Solution

Attempt 1: $P(X)=\binom{3}{1}\binom{1}{7}^1\binom{6}{7}^2= \frac{108}{343}$

Ans. 1.: (From textbook): $\frac{36}{343}$

Attempt 2: $P(X)=\binom{3}{1}\binom{1}{6}^1\binom{5}{6}^2= \frac{25}{72}$

 

[Ray Vickson]

Homework Statement

In order to highlight the problem I'm having here I have posted two questions, the answer for the second matches the textbook answer, the first does not. The wording of both questions appears to be the same. Have I gone wrong in the 1st question, or is the book incorrect?

Many thanks.

Q. 1. Three people were selected at random and asked on which day of the week their next birthday was falling. What is the probability that only one of the birthdays falls on a Sunday.

Q. 2. A fair die is thrown 3 times. Find the probability that there will be exactly one 6.

Homework Equations

The Attempt at a Solution

Attempt 1: $P(X)=\binom{3}{1}\binom{1}{7}^1\binom{6}{7}^2= \frac{108}{343}$

Ans. 1.: (From textbook): $\frac{36}{343}$

Attempt 2: $P(X)=\binom{3}{1}\binom{1}{6}^1\binom{5}{6}^2= \frac{25}{72}$

What is the *logic* you used in getting your answer? In other words, why do you write what you did write? (BTW: I get your answer.)

 

[odolwa99]

In Q.1 The arrangement is P(S, F, F) + P(F, S, F) + P(F, F, S). Where F is fail, i.e. not the selected day, so 6/7. And S is success, i.e. the selected day, so 1/7. Then multiply the success as shown in the 1st sentence and add the 3 totals for the answer.

The same logic applies with question 2, except now the odds are that F = 5/6 & S = 1/6.

For the books answer to be correct, factor only a successful day for one outcome, and ignore the remaining 2. I'm assuming that this is what the question is aiming for?

In the second question, 3 separate die rolls means that 3 separate probabilities will be accounted for, not just the 1st.

 

[haruspex]

You answers are correct. (Btw, it would be better not to use the same notation for both combinatorials and fractions. For the fractions use \frac{}{}.)

 

[odolwa99]

Ok, so the book is definitely wrong? Thanks.