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Anastaziya [24]
3 years ago
10

When rolling two six-sided number cubes, what is the probability of getting a sum of at least 5? 5/18 1/3 13/18 5/6

Mathematics
1 answer:
VashaNatasha [74]3 years ago
3 0

The sample space for two six sided cubes =

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)


Now the favourable events are those where the sum is greater than or equal to 5.

It means

(1,4) (1,5) (1,6)

(2,3) (2,4) (2,5) (2,6)

(3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)


It means

favourable events = 30

and total possible events = 36


So probability = 30/36   = 5/6

Hence 4th option is the correct one.

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<img src="https://tex.z-dn.net/?f=%2826%20%5Cdiv%20100%29%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5Ctimes%2010" id
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Answer:

\boxed{\bf \:  \cfrac{13}{5}}

<u>Or in Decimal:</u>

\boxed{\bf \: 2.6}

Step-by-step explanation:

<u>Given expression :-</u>

\sf \: ( 26 \div 100) \times 10

<u>Solution :-</u>

\sf  = (26 \div 100 )\times 10

This arithmetic expression may be rewritten as ;

\sf  =  \cfrac{26}{100}  \times 10

Step 1 : <u>Cancel the zero of 10 and one zero of 100</u> :-

\sf  =  \cfrac{26}{10 \cancel0}  \times 1 \cancel0

<em>Results to;</em>

\sf  =  \:  \cfrac{26}{10}  \times 1

\sf  =  \:  \cfrac{26}{10}

Step 2: <u>Cancel 26 and 10</u><u> </u><u>by 2</u> :-

\sf  =  \cfrac{ \cancel{26}}{ \cancel{10}}

<em>Results to;</em>

\sf = \cfrac{ \cancel{26} {}^{13} }{ \cancel{10} {}^{5} }

\sf  =  \cfrac{13}{5}

<em>It can also be in Decimal.</em>

That is;

\sf = 2.6

Hence, the answer of the expression would be 13/5 or 2.6 .

\rule{225pt}{2pt}

I hope this helps!

Let me know if you have any questions.

I am joyous to help!

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2 years ago
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