Deer ticks can be carriers of either Lyme disease or human granulocytic ehrlichiosis (HGE). Based on a recent study, suppose that 16% of all ticks in a certain location carry Lyme disease, 10% carry HGE, and 10% of the ticks that carry at least one of these diseases in fact carry both of them. If a randomly selected tick is found to have carried HGE, what is the probability that the selected tick is also a carrier of Lyme disease

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

10000

step-by-step explanation: divide by 2,