Respuesta :
If Daniel has two minutes and each game lasts for 15 seconds then he has eight tries to win. ( 120 seconds divided by 15 seconds = 8) And if his odds are .1 that means he has a 10% chance each game. 10% x 8 tries = 80% chance. or .8
Using the binomial probability concept, the probability of having a win right before the class starts is 0.383.
- Time taken to play a game = 15 seconds
- Number of games that can be played = (2 × 60) /15 = 8 games
Using the binomial probability relation :
- P(x = x) = nCx * p^x * q^(n-x)
- Probability of winning = 0.1
- n = number of trials = 8
Probability of having a win right before the class starts is the probability of having one win :
P(x = 1) = 8C1 × (0.1)¹ × (0.9)^7
P(x = 1) = 8 × 0.1 × 0.4782969
P(x = 1) = 0.383
Therefore, the probability of having a win right before the class starts is 0.383
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