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1 year ago

This probability distribution shows

Mathematics

1 answer:

Charra [1.4K]1 year ago

7 0

Answer:

The probability that a student earns a grade of A is 1/7.

Let E be an event and S be the sample space. The probability of E, denoted by P(E) could be computed as:

P(E) = n(E) / n(S)

As the total number of students = n(S) = 35

Students getting the grade A = n(E) = 5

So, the probability that a student earns a grade of A:

P(E) = n(E) / n(S)

= 5/35

= 1/7

Hence, the probability that a student earns a grade of A is 1/7.

Keywords: probability, sample space, event

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Rainbow [258]

Answer:

0.25% probability that they are both defective

Step-by-step explanation:

For each computer chip, there are only two possible outcomes. Either they are defective, or they are not. The probability of a computer chip being defective is independent of other chips. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

5% of the computer chips it makes are defective.

This means that $p = 0.05$

If an inspector chooses two computer chips randomly (meaning they are independent from each other), what is the probability that they are both defective?

This is P(X = 2) when n = 2. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,2}.(0.05)^{2}.(0.95)^{0} = 0.0025$

0.25% probability that they are both defective

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2 years ago

Steve and Ben have some money in their wallets. If Steve gives Ben $3, Ben will have twice as much as Steve. If Ben gives Steve

Zigmanuir [339]

Answer:

Money owned by Steve = $11

Money owned by Ben = $13

Step-by-step explanation:

Let x denotes the money owned by Steve and y denotes the money owned by Ben.

Then, Steve gives $3 to Ben

Money left with Steve = x-3 and Ben = y+3

Now, Ben will have twice as much as Steve.

⇒ y+3 = 2(x-3) .............(1)

If Ben gives Steve $7

Money left with Steve = x+7 and Ben = y-7

Then, the amount Ben has will be one-third that of Steve’s.

⇒ y-7 = $\frac{1}{3}$ (x+7) ..........(2)

Solving equation (1) and (2) by elimination method, we get

x = 11 and y = 13

⇒ Money owned by Steve = $11

Money owned by Ben = $13

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2 years ago

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Answer: See attached image.

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2 years ago

Solve by inspection: X+4=15

amid [387]

x + 4 = 15

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Step-by-step explanation:

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