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My name is Ann [436]
3 years ago
12

One of two small restaurants is chosen at random with equally likely probability, and then an employee is chosen at random from

the chosen restaurant. Restaurant #1 has 10 full-timers and 6 part-timers. Restaurant #2 has 7 full-timers and 9 part-timers. What is the probability that Restaurant #1 was chosen at random, given that a full-time employee was chosen? Your answers should be rounded to 4 digits after the decimal.
Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
8 0

Answer:

P(1 |F) = 10/17

Step-by-step explanation:

Let events

1 = restaurant 1

2 = restaurant 2

F = full-time worker chosen

P = part-time worken chosen

P(1 and F) = 1/2 * 10/16 = 5/16

P(2 and F) = 1/2 * 7/16 = 7/32

P( (1 or 2) and F ) = P(F) = 5/16+7/32 = 17/32

P(1 | F)          Probability of choosing restaurant 1 given a full-time was chosen

= P(1 and F) / P(F)

= 5/16  / (17/32)

= 5/16 * 32/17

= 10 / 17

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The best and most correct answer among the choices provided by your question is the fourth choice or letter D. The probability is 4/5.

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Hi there!

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