Question

A true–false question is to be posed to a husband and-wife team on a quiz show. Both the husband and the wife will independently give the correct answer with probability p. Which of the following is a better strategy for the couple?

(a) Choose one of them and let that person answer the question.

(b) Have them both consider the question, and then either give the common answer if they agree or, if they disagree, flip a coin to determine which answer to give

Short answer

Both strategies have probability $p$to succeed.

Wanted probability is that of answering correctly, divide it into events whose probability is given.

Step-by-step solution

Given information(part a)

True–false question is to be posed to a husband-and-wife team on a quiz show. Both the husband and the wife will independently give the correct answer with probability $p$.

Explanation(part a)

Events:

$\Omega -$sample space, the question is answered

$H-$event that the husband answers the question

$W-$event that the wife answers the question

$W_k-$event that the wife knows the answer to the question

$H_k-$event that the husband knows the answer to the question

$C-$event that the question is answered correctly

Probabilities:

$P\left(\Omega \right)=1$

$P\left(H_k\right)=p$

$P\left(W_k\right)=p$

$H_k$and $W_k$are independent of $H$and $W$, and of one another.

Step 3:Explanation(part a)

Strategy:

$H=W^c$

$P\left(H\right)=P\left(W\right)=\frac{1}{2}$

Event of answering correctly can occur in two ways: either wife knows and she is the one to answer or the husband knows and he answers

$P\left(C\right)=P\left(H_k\cap H\right)+P\left(W_k\cap W\right)$

$=P\left(H_k\right)·P\left(H\right)+P\left(W_k\right)·P\left(W\right)$

$=p·\frac{1}{2}+p·\frac{1}{2}$

$=p$

Final answer(part a)

$P\left(C\right)=$$p$

the strategy has probability $p$to succeed.

Wanted probability is that of answering correctly, divide it into events whose probability is given.

Step 5:Given information (part b)

A true–false question is to be posed to a husband-and-wife team on a quiz show. Both the husband and the wife will independently give the correct answer with probability p

Explanation(part b)

Events:

$\Omega$- sample space, the question is answered

$H$- event that the husband answers the question

$W$- event that the wife answers the question

$W_k$- event that the wife knows the answer to the question

$H_k$- event that the husband knows the answer to the question

$C$- event that the question is answered correctly

Probabilities:

$P\left(\Omega \right)=1$

$P\left(H_k\right)=p$

$P\left(W_k\right)=p$

$H_k$ and $W_k$ are independent of $H$ and $W$, and of one another.

Explanation(part b)

Strategy:

If they both agree that is their answer, if they do not, one of their answers is chosen at random.

Now event $C$happens if both agree and are correct, wife is correct and husband is not and the wife answers, or husband is correct and he answers.

Since no one changes opinion, when they agree let's say that half of the times the wife says the answer, then husband will answer

$P\left(C\right)=P\left(H_k{W}_k\right)+P\left(H_k{W}_k^{c}H\right)+P\left(H_k^c{W}_{k}W\right)$

$=P\left(H_k{W}_k\right)+P\left(H_k{W}_k^c\right)P\left(H\right)+P\left(H_k^c{W}_k\right)P\left(W\right)$

$=2·P\left(H_k{W}_k\right)\frac{1}{2}+P\left(H_k{W}_k^c\right)\frac{1}{2}+P\left(H_k^c{W}_k\right)\frac{1}{2}$

$=\frac{1}{2}\left[P\left(H_k{W}_k\right)+P\left(H_k{W}_k^c\right)\right]+\frac{1}{2}\left[P\left(H_k{W}_k\right)+P\left(H_k^c{W}_k\right)\right]$

$=\frac{1}{2}P\left(H_k\right)+\frac{1}{2}P\left(W_k\right)$

$=\frac{1}{2}p+\frac{1}{2}p$

$=p$

Final answer(part b)

$P\left(C\right)=$$p$

The strategy has probability$p$to succeed.

Wanted probability is that of answering correctly, divide it into events whose probability is given.