there are 52 cards in a deck, 12 of those cards are "face cards", so the remaining are number cards, namely 40.
52 = sample space
40 = favorable outcomes
P(number card | number card) = p(number) * p(number)
so the first time we pull one, there are 52 cards, the probability of a number card is 40/52, or 10/13, and we don't put it back in the deck.
the next time we pull another card, the cards are no longer 52 total, we pulled one out, they're only 51, namely 51 = sample space, and the number cards if we really pulled out before, are no longer 40, are 39, namely 39 = favorable outcomes.
probability of getting a number card the second time? 39/51 or 13/17.
[tex]\bf \stackrel{\textit{probability of getting a number card twice}}{\cfrac{10}{~~\begin{matrix} 13 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\cdot \cfrac{~~\begin{matrix} 13 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{17}\implies \cfrac{10}{17}~~\approx ~~ 0.59}~\hfill 59\%[/tex]
Answer:
105/221
Step-by-step explanation:
There are 52 cards in a deck.
Assuming 2-10 are the number cards
2,3,4,5,6,7,8,9,10 = 9
There are 4 suits
9*4 = 36 cards are number cards
P(1st card is a number card) = number card/ total
=36/52 = 9/13
We do not replace the card, so there are only 51 cards left, and only 35 number cards
P(2nd card is a number card) = number card/ total
=35/51
The probability of getting 2 number cards in a row is
P (number ,number) =P(1st card is a number)*P(2nd card is a number card)
= 9/13 * 35/51
Dividing the top and bottom by 3
= 3/13 * 35/17
=105/221
