Question

) Henry, Dina, Ray and Susan will be the only participants at a meeting. There
will be three soft chairs and one hard chair in the room where the meeting will
be held. No one can bring more chairs into the room. Henry and Ray will arrive
simultaneously, but Dina and Susan will arrive individually. The probability that
Dina will arrive first is 1
3
, and the probability that Susan will arrive first is 1
3
.
The probability that Dina will arrive last is 1
3
, and the probability that Susan
will arrive last is 1
3
. Upon arriving at the meeting, each of the participants will
select a soft chair, if one is available. If Henry and Ray arrive and see only one
unoccupied soft chair, they will flip a fair coin to determine who will sit in that
chair. By what percentage is the probability that Henry will sit in a soft chair
greater than the probability that Dina will sit in a soft chair? [15]

Answer 1

Answer:

25 %

Step-by-step explanation:

Probability Dina & Susan arrive first is 1/3

so Probability of Henry & Ray will arrive first is also 1/3

Probability Dina & Susan arrive last is 1/3

so Probability of Henry & Ray will arrive last is also 1/3

Person ho reaches last  has to sit on hard chair

Probability of Dina Sitting in Hard Chair = 1/3

Probability of Dina Sitting in Soft Chair = 1 - 1/3 = 2/3

Probability of Henry Sitting in hard Chair = (1/2)(1/3) = 1/6

Probability of Henry Sitting in Soft Chair = 1 - 1/6 = 5/6

Probability of Henry Sitting in Soft Chair - Probability of Dina Sitting in Soft Chair = 5/6 - 2/3  =  1/6

% probability that Henry will sit in a soft chair  greater than the probability that Dina  = ( (1/6)/(2/3) ) * 100  = 25 %