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Also if a die is rolled 4 times, what is the probability of 6 showing up at least once?

1) There are exactly 13 cards of any suit which means there are $13 \times 12 \times 11 \times 10 \times 9$ possible flushes and ${52 \choose 5}$ total possibilities which gave me .005 but the correct answer is .002.

2) The sample space is ${ 1-6,1-6,1-6,1-6}$ and the number of ways that any one number chosen in any column can be combined with any of the other numbers in the second column...etc. for ex: if we roll a 1 we have 6 possible pairings for the second roll {(1,1),(1,2),(1,3),(1,4)} then if we roll a 1 on the second roll we have {(1,1,1),(1,1,2),(1,1,3),(1,1,4),(1,1,5),(1,1,6)...I dont know how to model this. It seems to me their might be ${6 \choose 1} \times {6 choose 1} \times {6 \choose 1} \times {6 \ choose 1}$ and then where do I begin on the numerator or E?

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    If you have two separate questions, please ask them in two separate posts. – 2017-02-03

1 Answers 1

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a) You seek: the probability for selecting five from thirteen values, all of one from four suits, when selecting five from fifty-two cards. $$\dfrac{\binom{13}{5}\binom{4}{1}}{\binom {52}{4}}=\dfrac{\frac{13\cdot12\cdot 11\cdot 10\cdot ~9~}{~5~\cdot~4~\cdot~3~\cdot~2~\cdot~1}\cdot\frac{4}{1}}{\frac{52\cdot51\cdot 50\cdot 49\cdot 48~}{~5~\cdot~4~\cdot~3~\cdot~2~\cdot~1}}=\dfrac{33}{16660}\approx 0.002 $$

Remember, when the sample space is counted without replacement in any order, so too must the favoured space.

b) You seek: the probability for at least one from four die being a specified value and the other die each being one from five remaining values, when each of three die may independently select one from six values.   Alternatively the complement of the favoured event is for none of the die to be the specified value. $$\dfrac{\binom 51 5^3+\binom 42 5^2+\binom 43 5 + 1}{6 ^4}=\dfrac{6^4-5^4}{6^3}$$

When the sample space is counted with replacement in specific order, so too must the favoured space.