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Ann [662]
3 years ago
11

A math class has three girls and 7 boys in the 7th grade and five girls and five boys in the 8th grade the teacher randomly sele

cts a 7th grader and an 8th grader from the class 4A competition what is the probability that the students she select are boys write your answer as a fraction in simplest form
Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
7 0

Answer:

7/20

Step-by-step explanation:

Seventh grade: Total = 3 + 7 = 10

Number of boys = 7 boys

Probability of a boy being picked = 7/10

Eighth grade: Total 5 + 5 = 10

Number of boys = 5

Probability 5/10 = 1/2

The probability of the correct outcome = 7/10  *  1/2  = 7 / 20

Answer: 7 / 20

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A 52-card deck is thoroughly shuffled and you are dealt a hand of 13 cards. (a) If you have at least one ace, what is the probab
jasenka [17]

Answer:

a) 0.371

b) 0.561

Step-by-step explanation:

We can answer both questions using conditional probability.

(a) We need to calculate the probability of obtaining two aces given that you obtained at least one. Let's call <em>A</em> the random variable that determines how many Aces you have. A is a discrete variable that can take any integer value from 0 to 4. We need to calculate

P(A \geq 2 | A \geq 1) = P(A\geq 2 \cap A \geq 1) / P(A \geq 1)

Since having 2 or more aces implies having at least one, the event A \geq 2 \cap A \geq 1 is equal to the event A \geq 2. Therefore, we can rewrite the previous expression as follows

P(A \geq 2) / P(A \geq 1)

We can calculate each of the probabilities by substracting from one the probability of its complementary event, which  are easier to compute

P(A \geq 2) = 1 - P((A \geq 2)^c) = 1 - P((A = 0) \bigsqcup (A = 1)) = 1 - P(A = 0) - P (A = 1)

P (A \geq 1) = 1 - P ((A \geq 1)^c) = 1 - P(A = 0)

We have now to calculate P(A = 0) and P(A = 1).

For the event A = 0, we have to pick 13 cards and obtain no ace at all. Since there are 4 aces on the deck, we need to pick 13 cards from a specific group of 48. The total of favourable cases is equivalent to the ammount of subsets of 13 elements of a set of 48, in other words it is 48 \choose 13. The total of cases is 52 \choose 13. We obtain

P(A = 0) = {48 \choose 13}/{52 \choose 13} = \frac{48! * 39!}{52!*35!} \simeq 0.303  

For the event A = 1, we pick an Ace first, then we pick 12 cards that are no aces. Since we can pick from 4 aces, that would multiply the favourable cases by 4, so we conclude

P(A=1) = 4*{48 \choose 12}/{52 \choose 13} = \frac{4*13*48! * 39!}{52!*36!} \simeq 0.438      

Hence,  

1 - P(A = 1)-P(A=0) /1-P(A=1) = 1 - 0.438 - 0.303/1-0.303 = 0.371

We conclude that the probability of having two aces provided we have one is 0.371

b) For this problem, since we are guaranteed to obtain the ace of spades, we can concentrate on the other 12 cards instead. Those 12 cards have to contain at least one ace (other that the ace of spades).

We can interpret this problem as if we would have removed the ace of spades from the deck and we are dealt 12 cards instead of 13. We need at least one of the 3 remaining aces. We will use the random variable B defined by the amount of aces we have other that the ace of spades. We have to calculate the probability of B being greater or equal than 1. In order to calculate that we can compute the probability of the <em>complementary set</em> and substract that number from 1.

P(B \geq 1) = 1-P(B=0)

In order to calculate P(B=0), we consider the number of favourable cases in which we dont have aces. That number is equal to the amount of subsets of 12 elements from a set with 48 (the deck without aces). Then, the amount of favourable cases is 48 \choose 12. Without the ace of spades, we have 51 cards on the deck, therefore

P(B = 0) = {48 \choose 12} / {51 \choose 12} = \frac{48!*39!}{51!*36!} = 0.438

We can conclude

P(B \geq 1) = 1- 0.438 = 0.561

The probability to obtain at least 2 aces if we have the ace of spades is 0.561

4 0
3 years ago
Is this an example of a horizontal asymptote?
8_murik_8 [283]

Answer:

Step-by-step explanation:

Hi there,

The graph indicated is showing a horizontal asymptote. In fact, it is showing both a horizontal and a <em>vertical </em>asymptote.

To tell which type it is, notice where the graph "shoots off" and almost forms an imaginary straight line in one direction. Using this logic, the horizontal asymptote will be exactly horizontal, parallel to x-axis, and vertical asymptote will be exactly vertical, parallel to y-axis.

With this graph, we notice the horizontal asymptote is at y=0, where the x-axis is. The vertical asymptote is bit more difficult to determine graphically, but can definitely say it is past x=-10. We could determine it if we had the function, but that is not necessary for this question.

Study well, and persevere. If you liked this solution, leave a Thanks or give a rating!

thanks,

6 0
3 years ago
Can you pls help me i reallly need it
yan [13]

Answer:

D

Step-by-step explanation:

$49 one time payment so it is added.

$25 per month and he pays for a year(12 months) = 12 times 25

$49 + (12 × 25)

4 0
2 years ago
Help meee plssssssssssssssssssssssssssss
Drupady [299]

Answer:

either a or c no problem nskskdjdjd

3 0
2 years ago
What is the distance between the points 1 - 6 + -5 2
Nata [24]
(1, -6) and (-5, 2)

Distance between 1 and -5 = 6
Distance between -6 and 2 = 8

6²+8²=x² (PYTHAGOREAN THEOREM)
36+64=x²
x² = 100
x = 10

Note: Not -10 because <em>distance cannot be negative.</em>
5 0
3 years ago
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