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3 years ago

What is the probability of having an odd number in a single toss of a fair die

Mathematics

2 answers:

Cloud [144]3 years ago

5 0

The chance is 1/6 because it's one side out of 6 tries.

krok68 [10]3 years ago

3 0

6 numbers on a regular die, 1, 2,3 ,4 5, 6

3 are odd so odds are 3/6 reduced to 1/2

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Please help me ASAP! Thank you!

Maurinko [17]

In step 2, you are adding 35.10 and 84.60. You carry the 1 to the 1's place after you add up the tenths and the hundreds place.

6 0

4 years ago

Read 2 more answers

Use the commutative property of addition to simplify 12x - 36 - 6x.

nexus9112 [7]

$12x - 36 - 6x \ is \ simplified \ to \ 6(x-6)$

<h2>Explanation:</h2>

The commutative property of addition states that:

$a+b=b+a$

So applying this property to our expression we have:

$12x - 36 - 6x=-6x+12x-36 \\ \\ \\ So \ here: \\ \\ a=12x-36 \\ \\ b=-6x \\ \\ \\ Combining \ like \ terms: \\ \\ 12x - 36 - 6x=6x-36 \\ \\ \\ Common \ factor \ 6: \\ \\ 12x - 36 - 6x=6(x-6) \\ \\ \\ Finally: \\ \\ 12x - 36 - 6x \ is \ simplified \ to \ 6(x-6)$

<h2>Learn more:</h2>

Evaluating expression: brainly.com/question/14423698#

#LearnWithBrainly

6 0

3 years ago

This couch is $100.12 and I had $1,000 how much money do I have now

Zanzabum

1,000-100.12=899.88

you have $899.88

hope this helped

7 0

4 years ago

40 is what percent of 50

fredd [130]

40/50=0.8
0.8*100=80%
Hope this helps!

3 0

3 years ago

Read 2 more answers

A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade

Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

$P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}$

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

$P(x\geq1)=1-P(x$

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly n= 5

Number of blades in an assembly dull is M = 10.

The probability mass function is,

$P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}$

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

$P(x\geq 1)=1- P(x$

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly N = 5

Number of blades in an assembly dull is M = 10

From the information,

The probability that assembly is replaced (P) is 0.7069.

The probability that assembly is not replaced is (Q) is,

$q=1-p\\= 1-0.7069= 0.2931$

The geometric probability mass function is,

$P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)$

The probability that assembly is replaced no replaced until the third day of evaluation is,

$P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607$

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 2.

$P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005$

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 6.

$P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968$

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

$P(x\geq 1)=1- P(x$

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0

4 years ago