Question

The probabilities that a typist will make $0,1,2,3,4,5$ or more mistakes in typing a report are, respectively, $0.12,0.25,0.36,0.14,0.08,0.11$ TRUE/FALSE

Answer 1

Solution:

It is given that

P(0) = 0.12

Let P(n) represents the probability of the event that the typist will make in mistakes

It is given that,

$P\left(0\right)\mathrm{0.12.}P\left(1\right)=0.25,P\left(2\right)=0.36$

$P\left(3\right)=0.14,P\left(4\right)=0.08$

P (5 or more) 0.11

We know, the total probability of all the

outcomes of any event is 1.

Solution:

Now, $P\left(0\right)+P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)$

$+P\left(5ormore\right)$

$=0.12+0.25+0.36+0.14+0.08+0.11$

$<=1.06$

The obtained total probability is not equal to 1

So, the given statement is FALSE.