What is the probability of getting either a sum of 5 or at least one 5 in the roll of a pair of dice?

1 answer:

4 0

In a throw of 2 fair dice, there are 6*6=36 equiprobability outcomes.

To get a sum of 5, there are 4 ways, (1,4),(2,3),(3,2),(4,1) with probability of 4/36=1/9

To get at least one 5, there are 6+6-1=11 outcomes (note (5,5) has been counted in both, so subtracted from sum). The probability is 11/36

Since the two events are mutually exclusive (once we have a five, the sum can no longer be 5), we can add the probabilities to get the probability of one event or the other.

P(sum of 5 OR at least one 5)=1/9+11/36=4/36+11/36=15/36=5/9

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The simplified expression

Romashka-Z-Leto [24]

Answer:

$5x^2 y^2$

Step-by-step explanation:

We need to use the properties shown below to solve this:

1. $\sqrt[n]{x^a} =x^{\frac{a}{n}}$

2. $\sqrt{x}\sqrt{x} =x$

3. $\sqrt{x} \sqrt{y}=\sqrt{x*y}$

Area of a triangle is given by 1/2 * base * height, so we do that and simplify:

$A=\frac{1}{2}(\sqrt{5x^3} )(2\sqrt{5xy^4} )\\A=\frac{1}{2}(5x^3)^{\frac{1}{2}}*2*(5xy^4)^{\frac{1}{2}}\\A=\sqrt{5}x^{\frac{3}{2}}*\sqrt{5}\sqrt{x} } y^2\\A=\sqrt{5} \sqrt{5}x^{\frac{3}{2}} x^{\frac{1}{2}}y^2\\A=5*x^2y^2\\A=5x^2 y^2$