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nadya68 [22]
3 years ago
12

1. Using the formula for the probability of one event or another event, calculate the probability of drawing a card from a stand

ard deck that is either a jack or a club. Write out the formula and show your work. Write the final answer as a fraction in simplified form. (10 points)
Bonus question for extra credit: What is the probability of drawing two cards from a deck, without replacing the first card, and either getting two red cards or two aces? Hint: the probability of getting two aces is the probability of the first card being an ace, times the probability of the second card being a different ace. Show your work, and write your answer as a fraction in reduced form.
Mathematics
1 answer:
ddd [48]3 years ago
7 0
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
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The sum of 4 consecutive intergers is 50, what is the first interger in this relationship
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Answer:

x = 11

Step-by-step explanation:

Consecutive means in a row

Let x be the first integer

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Divide each side by 4

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\implies {\blue {\boxed {\boxed {\purple {\sf { \: ({ \: z - 2 \: })^{2}  \:   }}}}}}

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