A boat charter of St. Lucia takes tourists on a daily dolphin/whale watch cruise

Question

A boat charter of St. Lucia takes tourists on a daily dolphin/whale watch cruise. Its brochure claims an 81% chance of sighting a whale or a dolphin, and you can assume that sightings from day to day are independent. (a) If you take the dolphin/whale watch cruise on two consecutive days, what is the probability that you see a whale or a dolphin on both days? (b) If you take the dolphin/whale watch cruise on two consecutive days, what is the probability that you see a dolphin or a whale on at least one day? (c) If you want to have a 99% probability of seeing a whale or a dolphin at least once, what is the minimum number of days that you will need to take the cruise?

Explanation / Answer

Answer to part b)

Probability that a whale or dolphin is visible on a day is 0.81

We need to find the probability that if the curise is taken for two consecutive days, then the wale or dolphin is seen on atleast on of the two days

This means, it includes three cases:

Case I: The whale or dolphin is visible on day I and not visible on day II

P(case I) = 0.81 *(1-0.81) = 0.1539

.

Case II: The whale or dolphin is not visible on day I and is visible on day II

P(case II) = (1-0.81)*0.81 = 0.1539

.

Case III: when it says atleast one , this means the whale or dolphin can be visible on both the days as well

P(Case III) = 0.81*0.81 = 0.6561

.

Thus P(at least 1 day ) = P(Case I) + P(Case II) + P(Case III)

P(atleast 1 day) = 0.1539 + 0.1539 + 0.6561

P(atleast 1 day) = 0.9639.

Thus the probability of atleast one day is 0.9639

.

There is one more method to find this probability

P(at least 1 day) = 1 - P(none of the two days)

P(none of the two days) = (1-0.81)*(1-0.81)

P(none of the two days) = 0.0361

P(atleast 1 day) = 1 - 0.0361

P(atleast 1 day) = 0.9639

.

Answer to part c)

For P(atleast) = 1- P(none)

Let number of days be n

Thus ,

P(none) = (1-0.81)^n

P(none) = (0.19)^n

.

The question says that P(atleast) = 0.99

Thus :

0.99 = 1 - P(none)

0.99 = 1 - (0.19)^n

(0.19)^n = 0.01

n = ln(0.01) / ln(0.19)

n = 2.77

This implies minimun 3 days consecutive on cruise must be spent in order to be able to see the dolphin with 99% probability

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