Gabriel has determined that if he gets to the bus stop by 7: 30 am, there is a 0.15 probability that he misses the bus. if he is at the bus stop three consecutive days at 7: 30, what is the approximate probability that he catches the bus all three days?

0.61 probability Since there's a 0.15 probability of Gabriel failing to catch the bus each day, that means that there's a 1 - 0.15 = 0.85 probability of Gabriel successfully catching the bus each day. So the probability of Gabriel catching the bus all three days will be the probability of success raised to the 3rd power. So 0.85^3 = 0.614125 Rounding to 2 significant figures gives Gabriel only a 0.61 probability of catching the bus all three days.

The approximate probability that he catches the bus all three days is 0.61.

Step-by-step explanation:

Given:

Probability of missing the bus at 7:30 AM is, .

Let q be the probability of getting the bus at 7:30 AM. Then,

Then, probability of getting the bus all three days is,

Here, n is the number of days (here, d = 3 days)

Thus, the approximate probability that he catches the bus all three days is 0.61.

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1.75 km = 1750 m
1750 m - 800 m = 950 m
950 m + 910 m = 1860 m 
1917 m - 1860 m = 57 m

57 more meters