The number of wins and losses of two local basketball

Ответ:

csuggs8

10.03.2022

0.429

Step-by-step explanation:

From the given table, the total number of games were loss = 12+16=28

The number of games was played by the wolves and was a loss = 12

The total number of games played = 18+12+16+14=60

Now, the probability of games was played by the wolves and was a loss is given by ;-

$\text{P(wolf and loss)}=\dfrac{12}{60}=\dfrac{1}{5}$

The probability of games that were loss is given by :-

$\text{P(loss)}=\dfrac{28}{60}=\dfrac{7}{15}$

Now, the probability that a randomly selected game from the season was played by the wolves, given that it was a loss is given by :-

$=\dfrac{\text{P(wolf and loss)}}{\text{P(loss)}}\\\\=\dfrac{\dfrac{1}{5}}{\dfrac{7}{15}}\\\\\\=\dfrac{3}{7}=0.428571428571\approx0.429$

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