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

Their is not enough information

Step-by-step explanation:

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.

So

(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).

sec ф = 5/3 (quadrant 4)

let x = base

y = perp

z = hypotenuse

sec ф = hypotenuse/base = 5/3

⇒ z = 5 and x = 3

finding y:

z² = x² + y²

5² = 3² + y²

y² = 25-9

y² = 16

y = 4

now finding all other geometric functions of theta:

sin ф = y/z = 4/5

sin ф = -4/5 (sin is negative in the 4th quadrant)

cos ф = x/z = 3/5

cos ф = 3/5 (cos is positive in 4th quadrant)

tan ф = y/x = 4/3

tan ф = -4/3 (tan is negative in 4th quadrant)

cosec ф = z/y = 5/4

cosec ф = -5/4 (cosec is negative in the 4th quadrant)

cot ф = x/y = 3/4

tan ф = -3/4 (tan is negative in 4th quadrant)