Vtyjkyuk

On a game show, the -
contestants each day can win $1, 000, 000 by correctly guessing an integer between 1 and 100
inclusive (which is chosen randomly each day). Before guessing the contestant can ask one yes/no question of his or
her choice. Monday's contestant asked Is the number 57? and Tuesday's contestant asked ìs the number greater
than 50?. Let P (M) be the probability of Monday's contestant winning and let P(T) be the probability of Tuesday's
contestants winning (assume each contestant properly uses the information gained from the question). Which of the
following is true?
A. P (M)/
P (T)
<.1 B. 0.1 <
P (M)/
P (T)
< 0.9 C. 0.
9<
P (M)/
P (T)
< 1.1 D. 1.1 <
P (M)/
P (T)
< 2 E.
P (M)/
P (T)
< 2

If $W$ denotes the winning number then:$$P(M)=P(Mmid W=57)P(W=57)+P(Mmid Wneq57)P(Wneq57)=$$$$1cdotfrac1100+frac199cdotfrac99100=frac150$$

More directly you could say that Monday's contestant uses the possibility to ask two distinct numbers. If the first happens to be correct already then the asking of the second is virtual.

Further we also find:$$P(T)=frac150$$This because after answer "yes" as well as answer "no" there is a choice between $50$ equiprobable numbers.

edit (to give some clarity) concerning monday contestant.

The event $M$ (i.e. the monday contestant wins) can be split up in two mutually exclusive events:

the answer on his question is "yes" and (of course) his guess will be $57$ or the answer on his question is "no" and (of course) he guesses a number that differs from $57$ which appears to be the correct number.

In mathematical language this is: $$M=(McapW=57)cup(McapWneq57)$$

Note that $cap$ corresponds with "and" and $cup$ corresponds with "or".

If two events $A,B$ are mutually exclusive (i.e. they cannot happen both, or mathematically $Acap B=varnothing$ then we have the rule $P(Acup B)=P(A)+P(B)$ and applying that here we find:$$P(M)=P(McapW=57)+P(McapWneq57)$$Now it remains to find the two terms.

If $A,B$ are independent events (i.e. the probability that $A$ happens will not affect the probability that $B$ happens, and vice versa) then we have the rule:$$P(Acap B)=P(A)P(B)$$

However we cannot apply that here because we are definitely not dealing with events $M$ and $W=57$ that are independent here!

Then how to find $P(Acap B)$ if $A,B$ are events that are not independent? For that we use another rule that also results in a multiplication:$$P(Acap B)=P(Amid B)P(B)$$

Here $P(Amid B)$ stands for the probability that event $A$ happens under the condition that event $B$ happens.

Applying that on the first term we find: $$P(Mmid W=57)P(W=57)$$or a bit more shortly written:$$P(Mmid W=57)P(W=57)$$

Now what is the probability that monday wins if $57$ is the winning number? $1$ of course, because the answer on his question has been answered with "yes" and then he simply knows that $57$ is the winning number. He will surely guess that number and win. This tells us that $P(Mmid W=57)=1$. Further what is the probability that $57$ is indeed the winning number? $frac1100$ of course because there are $100$ equiprobable candidates. This tells us that $P(W=57)=frac1100$ and we can calculate the first terms as:$$P(Mmid W=57)P(W=57)=1cdotfrac1100=frac1100$$

Now it remains to find the second term which will have the looks:$$P(Mmid Wneq57)P(Wneq57)$$ What is the probability that monday wins if the winning number is not $57$? His question has been answered with "no" and this information tells him that the winning number must be one of the elements of $1,2,dots,56,58,dots,99,100$. They have equal chances so the conclusion is evidently that $P(Mmid Wneq57)=frac199$. But we must not forgot the other factor: $P(Wneq 57)$, i.e. the probability that the winning card is not $57$ or equivalently the probability that his question is answered by "no". For that we find quite easily $P(Wneq57)=1-P(W=57)=1-frac1100=frac99100$. Now we are equipped to find the second term: $$P(Mmid Wneq57)P(Wneq57)=frac199frac99100=frac1100$$This results in:$$P(M)=P(McapW=57)+P(McapWneq57)=frac1100+frac1100=frac150$$