Chances of winning a game are 60 %. If Anil has played the game 20 times, how many times can he expect to lose?

# Chances of winning a game are 60 %. If Anil has played the game 20 times, how many times can he expect to lose?

1. A
7
2. B
6
3. C
8
4. D
5

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### Solution:

It is given that the chances of winning a game are 60 %.
The probability states that there cannot be a 100 percent probability of something. And probability always stays between zero and hundred.
$\mathit{Probability of winning}=\frac{60}{100}$
$⇒\mathit{Probability of winning}=0.6$
Therefore, probability of losing a game is,
$\mathit{Probability of losing}=1-\mathit{Probability of winning}$
$⇒\mathit{Probability of losing}=1-0.6$
$⇒\mathit{Probability of losing}=0.4$
Anil played 20 games, so the number of times expected to lose $=0.4×20$
Number of times expected to lose $=8$
Thus, Anil expected to lose 8 games out of 20, if the chance of winning the game is 60%.
Hence, option 3 is correct.

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