Question

there are 6 people taking part in a raffle.
ann, bob, kira, lena, omar, and ravi.
suppose that prize winners are randomly selected from the 6 people.
compute the probability of each of the following events.
event a: the first four prize winners are bob, lena, ravi, and kira, regardless of order.
event b: bob is the first prize winner, lena is second, ravi is third, and kira is fourth.
write your answers as fractions in simplest form.
p(a) =
p(b) =

Explanation:

Step1: Calculate \( P(A) \)

Event A: The first four prize winners are Bob, Lena, Ravi, and Kira (regardless of order). The number of ways to choose 4 people out of 6 is given by the combination formula \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 6 \) and \( k = 4 \). The total number of ways to choose 4 people out of 6 is \( C(6, 4)=\frac{6!}{4!(6 - 4)!}=\frac{6!}{4!2!}=\frac{6\times5}{2\times1}=15 \). There is only 1 combination that includes Bob, Lena, Ravi, and Kira. So the number of favorable outcomes for Event A is 1. Thus, \( P(A)=\frac{1}{C(6, 4)}=\frac{1}{15} \)? Wait, no, wait. Wait, actually, when we are selecting the first four winners, the total number of possible ordered or unordered? Wait, no, in combinations, if we consider the number of ways to choose 4 people (unordered) from 6, the total number of possible 4 - person groups is \( C(6, 4) = 15 \), and there is 1 group that is {Bob, Lena, Ravi, Kira}. So the probability \( P(A)=\frac{\text{Number of favorable groups}}{\text{Total number of 4 - person groups}}=\frac{1}{15} \)? Wait, no, maybe I made a mistake. Wait, actually, the total number of ways to select 4 people out of 6 (as a group) is \( C(6, 4)=15 \), and the favorable group is 1, so \( P(A)=\frac{1}{15} \). Wait, but let's think again. Alternatively, if we consider permutations. Wait, the problem says "the first four prize winners are... regardless of order". So it's a combination problem. So total number of ways to choose 4 people from 6 is \( C(6, 4)=\frac{6!}{4!2!}=15 \), and the number of favorable combinations is 1 (the group {Bob, Lena, Ravi, Kira}), so \( P(A)=\frac{1}{15} \).

Step2: Calculate \( P(B) \)

Event B: Bob is first, Lena is second, Ravi is third, Kira is fourth. So we are looking for a specific permutation. The total number of ways to arrange 4 people (in order) from 6 is \( P(6, 4)=\frac{6!}{(6 - 4)!}=\frac{6!}{2!}=6\times5\times4\times3 = 360 \). But wait, actually, for the first position, we have 6 choices, second position 5 choices, third 4, fourth 3. But in Event B, the first is Bob (1 choice), second is Lena (1 choice), third is Ravi (1 choice), fourth is Kira (1 choice). So the number of favorable permutations is 1 (since we have a specific order: Bob, Lena, Ravi, Kira). The total number of ways to choose the first four winners in order is \( 6\times5\times4\times3=360 \)? Wait, no, the total number of possible ordered 4 - tuples (since order matters here, as the positions are first, second, third, fourth) from 6 people is \( P(6, 4)=6\times5\times4\times3 = 360 \). The number of favorable ordered 4 - tuples is 1 (only one way to have Bob first, Lena second, Ravi third, Kira fourth). So \( P(B)=\frac{1}{6\times5\times4\times3}=\frac{1}{360} \)? Wait, no, wait. Wait, the total number of possible outcomes for the first four positions: for the first position, 6 people, second 5, third 4, fourth 3. So total is \( 6\times5\times4\times3 = 360 \). The favorable outcome is 1 (Bob, Lena, Ravi, Kira in that order), so \( P(B)=\frac{1}{6\times5\times4\times3}=\frac{1}{360} \). Wait, but let's check again. For Event B, the order is fixed: Bob (1st), Lena (2nd), Ravi (3rd), Kira (4th). So the number of ways this can happen is 1 (since each position has only one specific person). The total number of ways to choose the first four winners in order is \( 6\times5\times4\times3 = 360 \), so \( P(B)=\frac{1}{360} \).

Wait, but maybe I made a mistake in Event A. Let's re - examine Event A. Event A is "the first four prize winners are Bob, Lena, Ravi, and Kira, regardless of order". So we are considering the set of the first four winners being exactly those four people, in any order. So the total number of possible sets of four people (since order doesn't matter for the group) is \( C(6, 4)=15 \), and there is 1 set that is {Bob, Lena, Ravi, Kira}, so \( P(A)=\frac{1}{15} \).

Answer:

\( P(A)=\frac{1}{15} \), \( P(B)=\frac{1}{360} \)