Question

There are people taking part in a raffle.
Bob, Hans, Jim, Kira, Lena, Omar, Ravi, and Soo.
Suppose that prize winners are randomly selected from the people.
Compute the probability of each of the following events.

Event A: Hans is the first prize winner, Ravi is second, and Bob is third.

Event B: The first three prize winners are Omar, Lena, and Jim, regardless of order.

Write your answers as fractions in simplest form.

p(a)=
p(b)=

Answer 1

Given:

8 people took part in a raffle

Event A : Hans is first prize ,  Ravi is second and Bob is third.

Event B: The first three prize winners are Omar , Lena and Jim regardless of order  

To Find:

Compute the probability of the two events A and B.

Solution:

Event A

In event A three positions are fixed .

Hans -1 Ravi -2 Bob -3

Therefore there is only 1 way to fill the 1st,2nd and 3rd position

Permutations can only be made among the other positions.

Positions -           ___  ___ ___ ___ ___ ___ ___ ___

Number of ways-  1        1      1       5     4     3      2      1

Total number of ways of arranging  8 people =   8 ×7×6×5×4×3×2×1

Number of ways to satisfy event A =1×1×1×5×4×3×2×1

P(A)=

=  $\frac{(5)(4)(3)(2)}{(8)(7)(6)(5)(4)(3)(2)(1)} \\\\\\frac{1}{336}$

P(B) -

First 3 prizes must be given to 3 persons regardless of order

Hence they can also be arranged in 3! ways.

Rest 5 can be arranged in 5! ways.

positions                   ___ ___ ___ ___ ___ ___ ___ ___

number of ways         3     2       1     5     4     3       2     1

total number of ways to arrange 8 people = 8!.

$P(B)=\frac{(3)(2)(1)(5)(4)(3)(2)(1)}{(8)(7)(6)(5)(4)(3)(2)}\\ P(B) =\frac{6}{(8)(7)(6)} \\P(B) =\frac{1}{56}$

P(A)= $\frac{1}{336}$        P(B) = $\frac{1}{56}$