Let's consider the following problems:

1. Man tosses the fair coin and throws the die. What is the probability he gets a head in the coin AND 5 or 6 in the die?

The original question is similar to this one. The results of these two actions (tossing the coin and throwing the die) are independent events. The winning scenarios are {H,5} and {H,6}. Note that {5,H} is basically the same as {H,5}, the coin landed on heads and die gave 5.

The probability would be: P(AandB)=P(A)*P(B)=

\frac{1}{2}*\frac{2}{6}=\frac{1}{6}

Answer: \frac{1}{6}

NOTE: Probability of A and B.

When two events are independent, the probability of both occurring is the product of the probabilities of the individual events: P(A and B) = P(A) x P(B)

2. Man tosses the fair coin and throws the die. What is the probability he gets a head in the coin OR 5 or 6 in the die?

Note this is not what the original question is asking.

In this case the probability would be: P(AorB)=P(A)+P(B)-P(AandB)=\frac{1}{2}+\frac{2}{6}-\frac{1}{2}*\frac{2}{6}=\frac{2}{3}

Answer: \frac{2}{3}

NOTE: Probability of A OR B.

If Events A and B are independent, the probability that either Event A or Event B occurs is:

P(A or B) = P(A) + P(B) - P(A and B)

When we say "A or B occurs" we include three possibilities:

A occurs and B does not occur;

B occurs and A does not occur;

Both A and B occur.

3. Man tosses the fair coin and throws the die for n times. What is the probability he gets a head in the coin AND 5 or 6 in the die for all n times?

P=(\frac{1}{6})^n

So, if it were: Man tosses the fair coin and throws the die for 3 times. What is the probability he gets a head in the coin AND 5 or 6 in the die 3 times?

Then simply: \frac{1}{6}*\frac{1}{6}*\frac{1}{6}=\frac{1}{216}

Answer: \frac{1}{216}

4. Man tosses the fair coin and throws the die for n times. What is the probability he gets a head in the coin AND 5 or 6 in the die k times? (Meaning that the success is when during one try (toss&throw) he gets heads and 5 or 6. So, k success tries out of n.)

Probability of success is \frac{1}{6} as we calculated in the example 1.

NOTE: If the probability of a certain event is p (1/6 in our case), then the probability of it occurring k times in n-time sequence is:

P = C^n_k*p^k*(1-p)^{n-k}

So if it were: Man tosses the fair coin and throws the die for 3 times. What is the probability he gets a head in the coin AND 5 or 6 in the die 2 times? (Meaning that the success is when during one try (toss&throw) he gets heads and 5 or 6. So, two success tries out of 3.)

Then: P =C^n_k*p^k*(1-p)^{n-k}= C^3_2*(\frac{1}{6})^2*(1-\frac{1}{6})^{(3-2)}=C^3_2*(\frac{1}{6})^2*\frac{5}{6}=\frac{5}{72}

Answer: \frac{5}{72}

5. Man tosses the fair coin and throws the die for n times. What is the probability he gets a head m times in the coin AND 5 or 6 in the die k times?

P(H)=\frac{1}{2} and Q(5or6)=\frac{1}{3}

The same as in the previous example: P = C^n_m*p^m*(1-p)^{n-m}*C^n_k*q^k*(1-q)^{n-k}=C^n_m*(\frac{1}{2})^m*(1-\frac{1}{2})^{n-m}*C^n_k*(\frac{1}{3})^k*(1-\frac{1}{3})^{n-k}=

=C^n_m*(\frac{1}{2})^m*(\frac{1}{2})^{n-m}*C^n_k*(\frac{1}{3})^k*(\frac{2}{3})^{n-k}

So if it were: Man tosses the fair coin and throws the die for 5 times. What is the probability he gets a head 2 times in the coin AND 5 or 6 in the die 4 times?

P = C^5_2*(\frac{1}{2})^2*(\frac{1}{2})^{(5-2)}*C^5_4*(\frac{1}{3})^4*(\frac{2}{3})^{(5-4)}=C^5_2*(\frac{1}{2})^2*(\frac{1}{2})^{3}*C^5_4*(\frac{1}{3})^4*(\frac{2}{3})=\frac{25}{1944}

6. A man alternately tosses a coin and throws a die beginning with the coin. Find the probability that he gets a head in the coin BEFORE 5 or 6 in the die.

The one case when {H,5} and {5,H} is relevant is the above example. And more I think about this problem more I think this is what was intended to ask in the question. Here, I understand why the word "alternately" is used and I understand why the stem is stating that coin is tossed first.

Basically we have the following case man is tossing the coin and throws the die for several times. We are asked to calculate the probability that he gets "heads" BEFORE he gets 5 or 6.

Though I haven't seen GMAT question of this type, but here is solution:

The wining scenarios would be:

{H, any value on die} meaning that during the first try he tossed and got the heads right away;

OR

{T, any value but 5 or 6},{H, any value on die} meaning he didn't got heads and 5 or 6 during the first try but got heads during the second.

...

The probability would be the sum of the probabilities of such scenarios:

{H,Any} + {T, not 5 or 6}{H, Any} + {T, not 5 or 6}{T, not 5 or 6}{H, Any} + {T, not 5 or 6}{T, not 5 or 6}{T, not 5 or 6}{H, Any} + {T, not 5 or 6}{T, not 5 or 6}{T, not 5 or 6}{T, not 5 or 6}{H, Any} +....=

=\{\frac{1}{2}*1\}+\{\frac{1}{2}*\frac{2}{3}\}*\{\frac{1}{2}*1\}+\{\frac{1}{2}*\frac{2}{3}\}\{\frac{1}{2}*\frac{2}{3}\}*\{\frac{1}{2}*1\}+\{\frac{1}{2}*\frac{2}{3}\}\{\frac{1}{2}*\frac{2}{3}\}\{\frac{1}{2}*\frac{2}{3}\}*\{\frac{1}{2}*1\}...+...=\frac{1}{2}+\frac{1}{2}*\frac{1}{3}+\frac{1}{2}*\frac{1}{3}*\frac{1}{3}+\frac{1}{2}*\frac{1}{3}*\frac{1}{3}*\frac{1}{3}+...

As you can see we have the geometric progression with first term b_1=\frac{1}{2} and common ratio r=\frac{1}{3}. When 0<common ration<1, then the sum of geometric progression equals to \frac{b_1}{1-r}.

Hence the above sum equals to \frac{\frac{1}{2}}{1-\frac{1}{3}}=\frac{\frac{1}{2}}{\frac{2}{3}}=\frac{3}{4}

Answer: \frac{3}{4}

Hope it's clear.

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