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If someone plays 4 games in a row what is the probability that they lose the first three games and win the 4 games ?

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To find the probability of losing the first three games and winning the fourth game, we need to know the probability of losing a single game and the probability of winning a single game.

Let's assume that the probability of losing a single game is \(p\) and the probability of winning a single game is \(q\). Since winning and losing are mutually exclusive events, \(p + q = 1\).

To calculate the probability of losing the first three games and winning the fourth game, we multiply the probabilities of each individual event together:

Probability = Probability of losing the first game * Probability of losing the second game * Probability of losing the third game * Probability of winning the fourth game

Since the events are independent, we can multiply the probabilities:

Probability = \(p \times p \times p \times q = p^3 \times q\)

So, the probability of losing the first three games and winning the fourth game is \(p^3 \times q\).

Please note that without specific values for \(p\) and \(q\), we cannot provide an exact numerical answer.
User Bob Moore
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