Please check my homework questions, are they correct? 
*

*If you toss 5 fair coins, in how many ways can you obtain at least one head?


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*$2^5 - 1 = 31$ ways -> correct?


*If you toss 6 fair coins, in how many ways can you obtain at least one head and one tail?


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*$2^6 - 2 = 62$ -> correct?


*If you roll two dice, how many ways can you obtain a different number on each die?


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*30 ways -> correct? 


*How many 3 digit counting numbers are not a multiple of 10?


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*$9 * 10 * 9 = 810$ -> correct?


*Among the 2598960 possible 5 card poker hands, how many contain at least one black card?


*

*$_{26}C_{5} = 65780$ -> correct?


*If license plates consist of 3 letters followed by 4 digits, how many different licenses can be created having at least one digit or letter repeated?


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*$26^3 * 10 ^ 4 = 175760000$ -> correct?


*Of the 60 people who went to the cash register 16 had blond hair and 19 had black hair. Determine the probability the next person to come to the register will have black hair?


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*P (black) $= \frac{19}{60}$ -> correct?


*Assume the probability is $\frac{1}{2}$ that a child born is a boy. If a family has three children. What is the probability that they have:


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*a) exactly one boy -> $\frac{3}{8}$

*b) at most two girls -> $\frac{7}{8}$



-> correct?


*

*Eight people (four married couples) arrange themselves randomly in eight consecutive seats in a row. Find the probability that the men will be in four adjacent seats?


*

*$\frac{4! * 6!}{8!} = \frac{3}{7}$ -> correct?



All help is truly appreciated. I just want to know if my answers are right. If not please let me know what I did wrong. Thanks!
 A: The first four look OK. For the fifth, you need to subtract $\binom{26}{5}$, the number of all red hands, from the total number of hands.
For the sixth, you must subtract the number with no repetitions from the total. The number with no repetitions is $(26)(25)(24)(10)(9)(8)(7)$. The total is $26^310^4$. So the answer is $26^310^4-(26)(25)(24)(10)(9)(8)(7)$. 
The seventh and eighth look right. The added question is done incorrectly. First of all there is the implicit assumption that there are $4$ men and $4$ women.  But even assuming that, $\frac{3}{7}$ is clearly much too large. 
My way of doing it is to observe that the seats for the men can be chosen in $\binom{8}{4}$ equally likely ways. Of these, only $5$ have the men next to each other. 
You could also use "permutations" to solve the problem. That multiplies both "top" and "bottom" by $(4!)^2$.
A: No. of ways of arranging these 8 persons randomly in those seats = 8!
Assuming that the four males are tied together as an object.
Then, there are only 5 objects (4 female-objects + 1 tied object) to be arranged.
The no. of ways of arranging these 5 objects = 5!
If the 4 males are now untied, the no. of ways that they can re-arrange among themselves = 4!
The probability of having the men sitting together in four adjacent seats = 5! * 4! / 8!
= 1 / 14
A: Way too late, but 3 isn't correct. You have to divide by two: $6C2 = 15$. E.g. 3 and 4 is the same as 4 and 3.
