Question

Three married couples have purchased theater tickets and are seated in a row consisting of just six seats. If they take their seats in a completely random fashion (randomorder), what is the probability that Jim and Paula (husband and wife) sit in the two seats on the far left? What is the probability that Jim and Paula end up sitting next to one another? What is the probability that at least one of the wives ends up sitting next to her husband?

Answer 1

Answer:

(a)  $\frac{1}{15}$

(b)   $\frac{1}{3}$

(c)   $\frac{2}{3}$

Step-by-step explanation:

(a)

what is the probability that Jim and Paula (husband and wife) sit in the two seats on the far left?

In the question given, it indicates that 3 married couples consists of 6 individual and more so there exists six seats.

Therefore, the total number of way for 6 individual to sit = 6! ways.

Jim and Paula = 2! ways

that Jim and Paula (husband and wife) sit in the two seats on the far left = (6! -2!) ways  = 4! ways

Now, to determine the probability that Jim and Paula sit in the seat on the far left, we have;

Probability(J&P) = $\frac{required outcome}{Total number of possible outcome}$

Probability(J&P)  =  $\frac{2!*4!}{6!}$

Probability(J&P)  = $\frac{(2*1)*4!}{6!}$

Probability(J&P)  = $\frac{(2*1)*4!}{6*5*4!}$

Probability(J&P)  = $\frac{2}{30}$

Probability(J&P)  = $\frac{1}{15}$

∴   the probability that Jim and Paula (husband and wife) sit in the two seats on the far left = $\frac{1}{15}$

(b)

What is the probability that Jim and Paula end up sitting next to one another?

The total number of way for 6 individual to sit = 6! ways.

Numbers of ways to seat next to one another are: (1-2, 2-3, 3-4, 4-5, 5-6) = 5! ways

combination of Jim and Paula = C₂

∴ the probability  that Jim and Paula will end up sitting next to one another will be:

P ( J&P) = $\frac{number of ways to seat}{Total number of seat__{C_{(J&P)}}}$

P ( J&P) = $\frac{5!}{6*5!C__2}$

P ( J&P) = $\frac{5!}{\frac{6*5!}{2*1} }$

P ( J&P) = $5!*\frac{2*1}{6*5!}$

P ( J&P) = $\frac{2}{6}$

P ( J&P) = $\frac{1}{3}$

∴   the probability that Jim and Paula end up sitting next to one another = $\frac{1}{3}$

(c)

What is the probability that at least one of the wives ends up sitting next to her husband?

The total number of way for 6 individual to sit = 6! ways.

To find the required outcome, let make this easier by:

denoting  the probability that at least one of the wives ends up sitting next to her husband with Pr (K).

Let $K_a$ be the event of the couple and as such the three couples will be (a=1,2,3), sitting next to each other.

Therefore,

Pr(K) = Pr[K₁ or K₂ or K₃]

Pr(K) = Pr[K₁ ∪ K₂ ∪ K₃]

This type of event is uniformly distributed, and as such! , we have:

Pr (K) = Pr(K₁) + Pr(K₂) + Pr(K₃) - Pr(K₁∩K₂) - Pr(K₂∩K₃) - Pr(K₁∩K₃) + Pr(K₁∩K₂∩K₃)

Pr (K) = 3 Pr($K_a$) - 3 Pr(K₁K₂) + Pr(K₁∩K₂∩K₃)  

($K_a$) represents a couple together with four others

i.e a couple together with four others = (1!+4!) = 5!

= 2 × 5! ways

K₁K₂ represents two couples together with two other couples

= 2²  for two couples with two other couples (4!)

= 2² × 4! ways

K₁∩K₂∩K₃ represent three couples together

= 2³  × 3! ways

Pr(K) =  $\frac{number of required outcome}{total possible outcomes}$

Pr(K) = $\frac{3(Pr(K_a)-3(Pr(K_1K_2)+Pr(K_1K_2K_3)}{6!}$

Pr(K) = $\frac{(3*2*5!)-(3*2^2*4!)+(8*3!)}{6!}$

Pr(K) = $\frac{(3*2*5!)-(3*4*4!)+(8*3!)}{6!}$

Pr(K) = $\frac{6(5*4)3!-(12*4)3!+(8)3!}{6*5*4*3!}$

Pr(K) = $\frac{6*5*4-12*4+8}{6*5*4}$

Pr(K) = $\frac{80}{120}$

Pr(K) = $\frac{8}{12}$

Pr(K) = $\frac{2}{3}$

∴  the probability that at least one of the wives ends up sitting next to her husband = $\frac{2}{3}$