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Sidana [21]
3 years ago
6

In the game of tic-tac-toe, if all moves are performed randomly the probability that the game will end in a draw is . Suppose si

x random games of tic-tac-toe are played. What is the probability that at least one of them will end in a draw?
Mathematics
1 answer:
algol133 years ago
7 0

Completed question:

In the game of tic-tac-toe, if all moves are performed randomly the probability that the game will end in a draw is 0.127. Suppose six random games of tic-tac-toe are played. What is the probability that at least one of them will end in a draw?

Answer:

0.557

Step-by-step explanation:

For each game, the probability of not end in a draw is 1 - 0.127 = 0.873. Thus, because each game is independent of each other, the probability of all of them not end in a draw is the multiplication of the probability of each one:

0.873x0.873x0.873x...x0.873 = 0.873⁶ = 0.443

Thus, the probability that at least one of them end in a draw is the total probability (1) less the probability that none of them en in a draw:

1 - 0.443

0.557

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Determine whether each expression is equivalent to 49^2t – 0.5.
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Answer:

None of the expression are equivalent to 49^{(2t - 0.5)}

Step-by-step explanation:

Given

49^{(2t - 0.5)}

Required

Find its equivalents

We start by expanding the given expression

49^{(2t - 0.5)}

Expand 49

(7^2)^{(2t - 0.5)}

7^2^{(2t - 0.5)}

Using laws of indices: (a^m)^n = a^{mn}

7^{(2*2t - 2*0.5)}

7^{(4t - 1)}

This implies that; each of the following options A,B and C must be equivalent to 49^{(2t - 0.5)} or alternatively, 7^{(4t - 1)}

A. \frac{7^{2t}}{49^{0.5}}

Using law of indices which states;

a^{mn} = (a^m)^n

Applying this law to the numerator; we have

\frac{(7^{2})^{t}}{49^{0.5}}

Expand expression in bracket

\frac{(7 * 7)^{t}}{49^{0.5}}

\frac{49^{t}}{49^{0.5}}

Also; Using law of indices which states;

\frac{a^{m}}{a^n} = a^{m-n}

\frac{49^{t}}{49^{0.5}} becomes

49^{t-0.5}}

This is not equivalent to 49^{(2t - 0.5)}

B. \frac{49^{2t}}{7^{0.5}}

Expand numerator

\frac{(7*7)^{2t}}{7^{0.5}}

\frac{(7^2)^{2t}}{7^{0.5}}

Using law of indices which states;

(a^m)^n = a^{mn}

Applying this law to the numerator; we have

\frac{7^{2*2t}}{7^{0.5}}

\frac{7^{4t}}{7^{0.5}}

Also; Using law of indices which states;

\frac{a^{m}}{a^n} = a^{m-n}

\frac{7^{4t}}{7^{0.5}} = 7^{4t - 0.5}

This is also not equivalent to 49^{(2t - 0.5)}

C. 7^{2t}\ *\ 49^{0.5}

7^{2t}\ *\ (7^2)^{0.5}

7^{2t}\ *\ 7^{2*0.5}

7^{2t}\ *\ 7^{1}

Using law of indices which states;

a^m*a^n = a^{m+n}

7^{2t+ 1}

This is also not equivalent to 49^{(2t - 0.5)}

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