Question

In a survey of 200 employees of a company, it was found that 89 like the lunch packs provided by
the Company while the rest did not. Out of these employees, if one employee is chosen at
random, what is the probability that the chosen employee;
i) Likes the lunch pack?
ii) Dislikes the lunch pack?

Answer 1

Step-by-step explanation:

1)89/200

like the lunch pack

2)111/200

dislike the lunch pack

Answer 2

The probabilities are i) 0.445      ii) 0.555

Given

• No. of Employees- 200
• Employees that like the lunch packs = 89

To Find

If one employee is chosen at random, what is the probability that the chosen employee

1. Like the lunch pack?
2. Dislikes the lunch pack?

Solution

Probability of a given event = number of possible outcomes/ total no. of outcomes

Let be A the event that an employee chosen at random likes the lunch pack

Hence, P(A)

= no. of employees that like the lunch pack/ total number f employees

=89/200

= 0.445

Probability of non-occurrence of any event = 1 - the probability of its occurrence

Hence the probability that the employee chose at random dislikes the lunch pack is

1 - P(A)

= 1 - 0.445

= 0.555

Hence the probabilities are

i) 0.445

ii) 0.555

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