Mathematics: Five cards are dealt from a standard 52-card deck. (a) What is the probability that we draw 1 ace, 1 two, 1 three, 1 four, and 1 five (this is one way to get a straight )? (Round your answer to five decimal places.) (b) What is the probability that we draw any straight (including straight flush and royal straight flush hands)? (Round your answer to four decimal places.)

Idrk but i think u should figure out the n dia n then the units it will make it way more easier...

Five cards are dealt from a standard 52-card deck.

Ответ:

rosie20052019

18.01.2022

Step-by-step explanation:

As there are total 52 cards in a deck and we have to draw a set of 5 cards, we can use the formula of combination to find the total number of possible ways of drawing 5 cards.

Number of ways to draw 5 cards = N_T

$N_T\;=\;({}^NC_k)\\\\N_T\;=\;({}^{52}C_5)\\\\N_T\;=\;2,598,960$

(a) Assuming the cards are drawn in order (would not affect the probability). The of getting Ace, 2, 3, 4 and 5 can be obtained by multiplying the probability of getting cards below 6 (20/52) with the probability of getting 5 different cards (4 choices for each card).

$P(a)\;=\;\frac{20}{52}*\frac{4}{52}*\frac{4}{51}*\frac{4}{50}*\frac{4}{49}*\frac{4}{48}\\\\P(a)\;=\; 1.3133*10^{-6}$

(b) For a straight we require our set to be in a sequence. The choices for lowest value card to produce a sequence are ace, 2, 3, 4, 5, 6, 7, 8, 9, or 10. Hence, the number of ways are $({}^{10}C_1)$ .

For each card we can draw from any of the 4 sets. It can be described mathematically as: $({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)\;=\;[({}^{4}C_1)^5]$

Therefore, the total outcomes for drawing straight are:

$N_S\;=\;({}^{4}C_1)*({}^{4}C_1)^5\;=\;10240$

Thus, the probability of getting a straight hand is:

$P(b)\;=\;\frac{N_S}{N_T}\\\\P(b)\;=\;\frac{10240}{2598960}\\\\P(b)\;=\; 0.0039$

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