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2 years ago

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Bill wants to see how many times he needs to roll a standard number cube before getting a 5. What is the probability that Bill w

ill roll a 5 on his third roll and not before?
a. 125/1296
b. 25/216
c. 1/6
d. 1/2

2 answers:

chubhunter [2.5K]2 years ago

5 0

5/6 • 5/6 • 1/6 = 25/216 the answer is b

Hoochie [10]2 years ago

4 0

Answer:

The answer is b. 25/216

Step-by-step explanation:

Bill wants to see how many times he needs to roll a standard number cube before getting a 5.

p = $\frac{1}{6}$ (on an average: roll 6 times before getting a 5)

The required probability can be found as :

$\frac{1}{5}(\frac{5}{6}* \frac{5}{6}* \frac{5}{6})$

= $\frac{5}{6}* \frac{5}{6}* \frac{1}{6}$

= $\frac{25}{216}$

So, the answer is b. 25/216

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kkurt [141]

Answer:

$(x- \frac{2}{5})^{2} = x^2 -\frac{4}{5}x + \frac{4}{25}$

Step-by-step explanation:

You have two methods to expand this binomial.

Method 1

If you have the expression:

$(x- \frac{2}{5})^{2}$

You can write the expression it in the following way:

$(x-\frac{2}{5})^{2}=(x-\frac{2}{5})(x-\frac{2}{5})$

Then, apply the distributive property:

$(x-\frac{2}{5})(x-\frac{2}{5}) = x^2 -\frac{2}{5}x -\frac{2}{5}x+ (\frac{2}{5})\frac{2}{5}$

Simplify the expression:

$(x-\frac{2}{5})^2= x^2 -\frac{4}{5}x+ (\frac{4}{25})$

---------------------------------------------------------------------------------------

Method 2

For any expression of the form:

$(a-b)^2$

Its expanded form will be:

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If

$a = x$

$b =\frac{2}{5}$

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This means that:

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