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3 years ago

Find each probability a coin landing heads up on every toss when it is tossed 3 times

Mathematics

2 answers:

inn [45]3 years ago

5 0

Answer:

Step-by-step explanation:Consider coin 1:

It has to be either HEADS or TAILS (obviously). If it's a “Fair” coin the probability of getting either a Heads or Tails is (1/2) for each option.

Conclusion(so far):

Prob (Heads Coin 1) = (1/2) AND

Prob (Tails Coin 1) = (1/2)

To meet the requirements of the question, the second coin toss must have the same result as for the first coin toss.

So if the result for Coin 1 was Heads then the result for Coin 2 must also be Heads.

(Prob (Heads Coin 1 AND Heads Coin 2) = (Prob Heads Coin1) * (Prob Heads Coin 2) = (1/2)*(1/2) = (1/4)

Conclusion (so far) (2 coins tossed):

Prob of two coin tosses BOTH giving Heads is (1/2)*(1/2) = (1/4)

AND similarly

Prob of two coin tosses BOTH giving Tails is also (1/2)*(1/2) = (1/4)

To meet the requirements of the question, the third coin toss will, by definition be either HEADS or TAILS.

AND

Prob (Coin 3 Heads) = (1/2) AND

Prob (Coin 3 Tails) = (1/2)

The question is clear: We are asked to find the Probability that the (3 tosses) are the same.

That is, if the first toss (Coin 1) is Heads, then the results for Coin 2 and Coin 3 tosses must also be Heads.

Similarly if the the first toss (Coin 1) is Tails, then the results for Coin 2 and Coin 3 tosses must also be Tails.

So for the tosses for Coins 1, 2 and 3 to each be Heads, we have:

Prob (Coin 1 Heads, Coin 2 Heads, Coin 3 Heads) = (1/2)*(1/2)*(1/2) = (1/8)

AND

Prob (Coin 1 Tails, Coin 2 Tails, Coin 3 Tails) = (1/2)*(1/2)*(1/2) = (1/8)

There are NO OTHER NUMBER COMBINATIONS WHICH MEET THE REQUIREMENTS OF THE QUESTION other than 3 Heads AND 3 Tails.

It is important to note THAT THE SITUATIONS ABOVE (3 Heads AND 3 Tails) are BOTH VALID SOLUTIONS.

Therefore

The Probability for 3 Heads is (1/8)

AND

The Probability for 3 Tails is (1/8)

BUT BOTH THESE SOLUTIONS ARE VALID PROBABILITIES and BOTH meet the requirements of the question.

Therefore the overall probability is THE SUM OF BOTH PROBABILITIES, that is:

The probability that after 3 tosses, the tosses of each coin ALL give the same result is:

(1/8) + (1/8) = (2/8) = (1/4) or 0.25 or 25%

This means that after 3 coin tosses there is a 25% probability that we will get either 3 Heads or 3 Tails.

(A note of interest: It would be an error to ignore the fact that there are TWO valid number combinations which meet the requirements of the question, not one).

(In my analysis above I have identified ONLY the valid-number combinations and spent little time considering non-valid-number combination).

MrRissso [65]3 years ago

3 0

Answer:

Their is not enough information

Step-by-step explanation:

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Vesna [10]

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The answer is "152 ft"

Step-by-step explanation:

Please find the attached file.

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2 years ago

X²+y²-6x+14y-1=0 and please show your work so I can learn

Eva8 [605]

Hello there,

I hope you and your family are staying safe and healthy during this winter season.

$x^2 + y^2 -6x+14y-1=0$

We need to use the Quadratic Formula*

$x =\frac{-b+\sqrt{b^2}-4ac }{2a}$ , $\frac{-b-\sqrt{b^2} -4ac }{2a}$

Thus, given the problem:

$a = 1, b=-6, c=y^2+14y-1$

So now we just need to plug them in the Quadratic Formula*

$x=\frac{6+2\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$ , $x=\frac{6-\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$

As you can see, it is a mess right now. Therefore, we need to simplify it

$x=\frac{6+2\sqrt{10-y^2-14y} }{2}$ , $x = \frac{6-2\sqrt{10-y^2-14y} }{2}$

Now that's get us to the final solution:

$x=3+\sqrt{10-y^2-14y}$ , $x=3-\sqrt{10-y^2-14y}$

It is my pleasure to help students like you! If you have additional questions, please let me know.

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2 years ago

A collection of 5 positive integers has mean 4.4, unique mode 3 and median 4. If an 8 is added to the collection, what is the ne

Sphinxa [80]

9514 1404 393

Answer:

4.5

Step-by-step explanation:

It appears the 5 integers must be {3, 3, 4, 5, 7}.

For the mode to be unique, there must be more than one '3', and there cannot be any other repeated numbers. For 4 to be the median, it must be the 3rd-highest number. Then the remaining numbers must be integers greater than 4 with a sum of 12 for the mean to be 4.4.

When 8 is added to the collection, the median is now the average of 4 and 5, so is 4.5.

The new median is 4.5.

_____

Comment on vocabulary

mean: the sum of numbers divided by the number of numbers. 5 numbers with a mean of 4.4 will have a sum of 5×4.4 = 22.

median: the middle number of a collection of an odd number of numbers, when sorted from lowest to highest. It is the 3rd number of a group of 5 numbers. When there are an even number of numbers in the group, the median is the average of the middle two in the sorted list. Here, that means the 3rd number in the list of 5 is '4'. When 8 is added, 4 becomes one of the numbers contributing to the average that will be the new median.

mode: the member of the group that appears the most times. When the mode is unique, it means there are no other numbers repeated as often. Here, we know the 3rd number (of 5) is 4, so there can only be two numbers '3' in the collection. That 3 is the unique mode means no other numbers are repeated at all.

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3 years ago

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Stolb23 [73]

Answer:

can you elaborate?

Step-by-step explanation:

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3 years ago

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