Medical analysts predict that one-third of all people who are infected

Question

Tough and Tricky questions: Probability.

Medical analysts predict that one-third of all people who are infected by a certain biological agent could be expected to be killed for each day that passes during which they have not received an antidote. What fraction of a group of 1,000 people could be expected to be killed if infected and not treated for three full days?

A) 16/81
B) 8/27
C) 2/3
D) 19/27
E) 65/81

A) 16/81
B) 8/27
C) 2/3
D) 19/27
E) 65/81

Kudos for a correct  solution .

Source: Chili Hot GMAT

omegacube · Answer

at the end of day 1 - Dead = 1/3, alive = 1 - 1/3 = 2/3
at the end of day 2 - Dead = 1/3 + (1/3).(2/3), alive = 1 - {1/3 + (1/3).(2/3)} = 4/9
at the end of day 3 - Dead = 1/3 + (1/3).(2/3) + (1/3).(4/9) = 19/27

Hence D

Van4ez · Answer

At the end of each day we will have 2/3 alive. So after 3 days we will have (2/3)^3 people alive. Therefore, fraction of dead people will be 1-(2/3)^3=19/27.
The correct answer is D.

arunspanda · Answer

Medical analysts predict that one-third of all people who are infected by a certain biological agent could be expected to be killed for each day that passes during which they have not received an antidote. What fraction of a group of 1,000 people could be expected to be killed if infected and not treated for three full days?

A) 16/81
B) 8/27
C) 2/3
D) 19/27
E) 65/81

A) 16/81
B) 8/27
C) 2/3
D) 19/27
E) 65/81

People expected to be killed at the end of first day =  1000*(\frac{1}{3}) 
Remaining people, expected not to be killed at the end of first day=  1000*(\frac{2}{3})

People expected to be killed at the end of second day =  1000*(\frac{2}{3})(\frac{1}{3}) 
Remaining people, expected not to be killed at the end of second day=  1000*(\frac{2}{3})^2

One may observe that the number of people expected as not killed follows a geometric progression with initial value as 1000 and fixed common ratio ( \frac{2}{3} ). Thus, people expected not to be killed at the end of third day =  1000*(\frac{2}{3})^3 = (\frac{8}{27})*1000

Therefore, fraction of 1,000 people that could be expected to be killed in three full days =  1-\frac{8}{27}= \frac{19}{27}

Answer:  D

KarishmaB · Answer

Since we need the answer in fraction, we can ignore the 1000.

At the end of every day, 2/3rd of the previous day's number is left.

So at the end of day three, (2/3)*(2/3)*(2/3) = 8/27 are left

So by the end of day three, 1 - 8/27 = 19/27 could be expected to be killed.

Answer (D)

ScottTargetTestPrep · Answer

Instead of using 1,000, let’s use variable t.

After day 1, (1/3)t have not survived and (2/3)t are left.

After day 2, (2/3)t x 1/3 = (2/9)t have not survived and (2/3)t x 2/3 = (4/9)t are left.

After day 3, (4/9)t x 1/3 = (4/27)t have not survived and (4/9)t x 2/3 = (8/27)t are left.

Thus, after 3 days the fraction that have not survived is (1/3)t + (2/9)t + (4/27)t = (9/27)t + (6/27)t + (4/27)t = (19/27)t.

Answer: D

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Medical analysts predict that one-third of all people who are infected

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Medical analysts predict that one-third of all people who are infected

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards