Survival Probability Of The 6th Fly that Attempt To Pass A Spider

Problem
A spider eats three flies a day. Until he fills his quota, he has an even chance of catching any fly that attempts to pass. A fly is about to make the attempt. What are the chances of survival, given that five flies have already made the attempt today?

A. 1/2	C. 3/4
B. 1/4	D. 2/3

Answer Key

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[ C. 3/4 ]

Solution

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$P_\text{survive} = 1 - P_\text{capture}$

The 1st three flies completed the quota (the 1st three flies were captured)
$Q_1 = (1/2)^3 = 1/8$

The 4th fly completed the quota (only 2 were captured from the 1st three flies)
$Q_2 = (1/2)^2 (1/2)^1 \cdot (3C2) \times (1/2) = 3/16$

The 5th fly completed the quota (only 2 were captured from the 1st four flies)
$Q_2 = (1/2)^2 (1/2)^2 \cdot (4C2) \times (1/2) = 3/16$

Q_quota = Probability that the spider made its quota of three flies before the 6th fly made the attempt to pass:
$Q_\text{quota} = Q_1 + Q_2 + Q_3 = 1/8 + 3/16 + 3/16$

$Q_\text{quota} = 1/2$ ← this is the probability that the spider will not attack the 6th fly

P_attack = Probability that the spider will attack the 6th fly
$P_\text{attack} = 1 - Q_\text{quota} = 1 - 1/2$

$P_\text{attack} = 1/2$

P_capture = Probability that the 6th fly will get caught by the spider
$P_\text{capture} = P_\text{attack} \times P_\text{catch a fly} = 1/2 \times 1/2$

$P_\text{capture} = 1/4$

Hence,
$P_\text{survive} = 1 - 1/4$

$P_\text{survive} = 3/4$ ← answer

Survival Probability Of The 6th Fly that Attempt To Pass A Spider

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