Answer:
p=0.25
Step-by-step explanation:
Given that a club can select one member to attend a conference. All of the club officers want to attend. There are a total of four officers, and their designated positions within the club are President (P), Vice dash President (Upper V )comma Secretary (Upper S )comma nbspand Treasurer (Upper T ).
Sample space would be
a){ {P}, {V}, {S} {T}} is the sample space with notations standing for as given in the question
b) Each sample is equally likely. Hence we have equal chances for selecting any one out of the four.
If probability of selecting a particular sample of size I is p, the by total probability axiom we have
\begin{gathered}4p =1\\p =0.25\end{gathered}
4p=1
ab=ab=ab 3679y
Step-by-step explanation:
after all y quo hsubeu
Answer:
0.2364
Step-by-step explanation:
We will take
Lyme = L
HGE = H
P(L) = 16% = 0.16
P(H) = 10% = 0.10
P(L ∩ H) = 0.10 x p(L U H)
Using the addition theorem
P(L U H) = p(L) + P(H) - P(L ∩ H)
P(L U H) = 0.16 + 0.10 - 0.10 * p(L u H)
P(L U H) = 0.26 - 0.10p(L u H)
We collect like terms
P(L U H) + 0.10P(L U H) = 0.26
This can be rewritten as:
P(L U H)[1 +0.1] = 0.26
Then we have,
1.1p(L U H) = 0.26
We divide through by 1.1
P(L U H) = 0.26/1.1
= 0.2364
Therefore
P(L ∩ H) = 0.10 x 0.2364
The probability of tick also carrying lyme disease
P(L|H) = p(L ∩ H)/P(H)
= 0.1x0.2364/0.1
= 0.2364
