PLEASE HELP ME THIS IS URGENT I AM BEING TIMED!!A game has 150 cards, and five of them are bonus cards.

6 answers
Question:

PLEASE HELP ME THIS IS URGENT I AM BEING TIMED!! A game has 150 cards, and five of them are bonus cards. If all the cards are on the table, what is the probability of getting a bonus card when choosing a card at random? Write your answer as a fraction.

Answers

1/30

Step-by-step explanation:

well if there are 150 cards on the table and 5 are bonus, this can be turned into a simple fraction 5/150. just simplify and you are good to go!

1/30 or 3.333%

Step-by-step explanation:

11/19

Step-by-step explanation:

If the probability is 8/19, the reverse is 11/19. That is the odds.

Hope this finds you well on a cold day!  

a) 1:12

b) 2:10, 1:5

c) I'm not really sure what to put here, basically Jenna is lucky and in 10 turns got bonus points twice when the chances are 1 out of 12

d) 20:100

Different scenarios

H - 1

H - 2

H - 3

H - 4

H - 5

H - 6

T - 1

T - 2

T - 3

T - 4

T - 5

T - 6

He expects that his bonus will be $210,000.

Step-by-step explanation:

For each plate appearence, there are only two possible outcomes. Either he gets on base(base hit or bases-on-balls), or he does not. The probability of getting on base on each plate appearence is independent of other plate appearences. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

600 plate appearances this year

So n = 600.

Expected number of hits

[tex]p = \frac{1}{3}[/tex]

So

[tex]E(X) = np = 600\frac{1}{3} = 200[/tex]

Expected number of base on balls.

[tex]p = \frac{1}{6}[/tex]

So

[tex]E(X) = np = 600\frac{1}{6} = 100[/tex]

Bonus

$1,000 for each hit and $100 for each base-on-balls he gets.

200*1,000 + 100*100 = 210,000

He expects that his bonus will be $210,000.

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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