# 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**

**Solution**

$Q_1 = (1/2)^3 = 1/8$

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*