Step-by-step explanation:
This is a probability related question, let the event be E
We know that the likelihood of an event happening is given as
Pr(E)=1
if an event will not occur the probability is
Pr(E)=0
a. This event is impossible: Pr(E)=0
b.This event will occur more often than not, but is not extremely likely:
Pr(E)=0<E>0.5
c.This event is extremely unlikely, but it will occur once in a while in a long sequence of trials:
Pr(E)=0<E<0.5
d.This event will occur for sure: Pr(E)=0
Answer:
as in pic below
Step-by-step explanation:
Answer:
<h2>the c one</h2>
Step-by-step explanation:
<h2>hope it will help</h2>
Just try to follow my description. When two persons are in the cabin, there is only 1 handshake. When a third person comes, he will have to handshake the two people who came before him. So, there will be 2 handshakes. When a fourth person comes, he would make 3 handshakes with the 3 people who came before him. When the fifth person comes, he would make 4 handshakes with the 4 people who came before him. So, you see there is a pattern. The number of handshakes is 1 less than the total number of people inside the cabin. So, if there are 14 people in the cabin, the last person to come in would have to make 13 handshakes with the 13 people who came in first. Obtaining the sum of all the handshakes starting from the 2 handshakes initially to the 13 handshakes, the sum would be
Total handshakes = 2+3+4+5+6+7+8+9+10+11+12+13
Total handshakes = 90
Therefore, there will be a total of 90 handshakes made within the cabin of 14 people.
what a mess?!
please write your question better...
if you don't want the wrong answer
Step-by-step explanation:
what should we do?
