Ingredients please.ingredients are not mentioned
This problem is a combination of the Poisson distribution and binomial distribution.
First, we need to find the probability of a single student sending less than 6 messages in a day, i.e.
P(X<6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)
=0.006738+0.033690+0.084224+0.140374+0.175467+0.175467
= 0.615961
For ALL 20 students to send less than 6 messages, the probability is
P=C(20,20)*0.615961^20*(1-0.615961)^0
=6.18101*10^(-5) or approximately
=0.00006181
Answer:
The answer is
Step-by-step explanation:
Step by step
Let <em>x</em> = 8.132132132...
Then 1000<em>x</em> = 8132.132132132...
Subtract <em>x</em> from 1000<em>x</em> to eliminate the fractional part:
1000<em>x</em> - <em>x</em> = 8132.132132132... - 8.132132132...
999<em>x</em> = 8132 - 8 = 8124
<em>x</em> = 8124/999 = 2708/333
