i guess we'll never know *dance montage*
Answer:
The probability that he ends up with a full house is 0.0083.
Step-by-step explanation:
We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.
We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).
We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);
- If he is given with two kings.
- If he is given one king and one ace.
Only in these two situations, he will end up with a full house.
Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.
So, the ways in which we can draw two kings from available three kings is given by = 
{∵ one king is already there}
= 
{∵ 
}
= 
= 0.0028
Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = 
= 
= 
= 0.0055
Now, probability that he ends up with a full house = 
= 
= <u>0.0083</u>.
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:
n!/((n-r)!r!)
In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.
5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5
5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10
5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10
5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5
Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.
Now we add together the combinations
1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.
The answer is 32.
-------------------------------
Note: There is also an equation for permutations which is:
n!/(n-r)!
Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.
We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.
I hope this helps! If you have any questions, let me know :)
Answer:
89
Step-by-step explanation:
i think
Answer:
960
Step-by-step explanation:
9600/10∧1=960
