A standard deck has 52 cards.
A standard deck has 4 jacks.
A standard deck has 13 clubs.
From this, we can derive the following:
The probability of drawing a jack is 4/52 or 1/13
The probability of drawing a club is 13/52 or 1/4
But since the problem asks for drawing jack or club, therefore we should add the 2 probabilities, making 17/52. This is not the final answer yet. We know that there is a jack of clubs, therefore we need to subtract 1 from the probabilities since jack of clubs were considered in the 2 categories of probability.
With that being said, the probability of drawing a club or a jack is 16/52 or 4/13
Bonus Question:
The first thing you need to do here is find the probability of each scenario. First let's do what is given, the probability of drawing 2 aces. Since there are 4 aces in a deck of 52, we can easily say that the probability of drawing an ace is 4/52. However for our second draw, the probability of drawing a different ace is 3/51. This is so since we already drew a card that is an ace, hence we need to subtract one from the total aces (4-1) and from the total cards in the deck (52-1). In getting the probability of drawing two aces, we need to multiply the said probabilities: 4/52 and 3/51, resulting to 1/221.
For the second scenario, the drawing of 2 red cards, we just use the same concept but in this, we are already considering the 2 red cards in the first scenario, therefore the chance of drawing a red on our first draw is 24/52. For our second, we just need to subtract one card, therefore 23/51. Multiply these two and we will get 46/221.
Now, the problem asks for the chance of drawing either 2 reds or 2 aces, therefore we add the probabilities of the 2 scenarios:
46/222 + 1/221 = 47/221
Summary:
First Scenario:
4/52 + 3/51 = 1/221
Second Scenario:
24/52 + 23/51 = 46/221
Chances of drawing 2 red cards or 2 aces:
1/221 + 46/221 = 47/221
Answer:
5 over 3 is the slope
(0,4/3) is the y- intercept
Step-by-step explanation:
Answer:
13
Step-by-step explanation:
3n+1
Let n= 4
3(4)+1
12+1
13
2/3 is left over
I hope this is the answer you are looking for.
The Answer Is C. 64, hope I helped
