<u>Answer:</u>
Probability that none of the meals will exceed the cost covered by your company=0.2637
Probability that one of the meals will exceed the cost covered by your company=0.4945
Probability that two of the meals will exceed the cost covered by your company=0.2197
Probability that three of the meals will exceed the cost covered by your company=0.02197
<u>Step-by-step explanation:</u>
15 restaurants located in Boston is relevant for this question.
One third of 15 means, 5 restaurants will always exceed 50$
So,
We can use combinations, as we don’t know the exact 5 restaurants out of the 15
None of the meals will exceed,
10 restaurants will not exceed the rate and we will surely have to go to 3 restaurants so
10C3/15C3
= (10*9*8/1*2*3)/ (15*14*13/1*2*3)
=120/455
=0.2637
One of the meals will exceed,
We have to visit 3 restaurants but one will exceed.
So,
(10C2*5C1)/15C3
= ((10*9/1*2) *5)/ (15*14*13/1*2*3)
=45*5/455
=225/455
=0.4945
Two of the meals will exceed,
We visit 3 restaurants but two will exceed.
So,
(10C1*5C2)/15C3
= ((10*5*4/1*2))/ (15*14*13/1*2*3)
=100/455
=0.2197
All 3 meals will exceed.
So,
5C3/15C3
= (5*4*3/1*2*3)/ (15*14*13/1*2*3)
=10/455
=0.02197
