Probability of Being Bitten by a Snake Problem

I'm having trouble with the following problem:


The problem looked easy enough, until I realized that my limited logic did not cover all possibilities, so I haven't been able to solve it.

This is one my attempts: *I'm not sure how to incorporate the fact that the first participate may die after the first snake, or the second/third snake.



Any steps in the right direction would be very helpful, thanks!

Initially there are 100 snakes of whom 10 are "bad". The probability the first person gets bit by the first snake is 10/100= 1/10. In that case, there are 99 snakes left, 9 of them "bad", 90 "good". The probability the second person will get three good "good" snakes, and so not get bitten, is (90/99)(89/98)(88/97). The probabilty of this happening is (1/10)(90/99)(89/98)(88/97).

The probability the first person will not be bit by the first snake but will be bit by the second is (90/100)(10/99)= 1/11. In that case, there are 98 snakes left, 9 of them "bad", 89 "good". The probability the second person will get three "good" snakes, and so not get bitten, is (89/98)(88/97)(87/96). The probability of that happening is (1/11)(89/98)(88/97)(97/96).

Do you see the idea? Do the same for the first person getting two "good" snakes and then a "bad" snake and for the first person getting three "good" snakes. The probability the second person will NOT be bitten is the sum of those four numbers.

Sssssnakes are always evil ....... never good within 10 ft of my leg....

Mathlete said:

I'm having trouble with the following problem:
View attachment 2716

The problem looked easy enough, until I realized that my limited logic did not cover all possibilities, so I haven't been able to solve it.

This is one my attempts: *I'm not sure how to incorporate the fact that the first participate may die after the first snake, or the second/third snake.

View attachment 2717

Any steps in the right direction would be very helpful, thanks!

Click to expand...

It doesn't actually say the bites are fatal. I worked it through assuming that person 1 chose three snakes - and wound up with four cases: either 0, 1, 2, or 3 "bad" snakes have been removed from the basket when person 2 starts choosing. Then I continued each of the four cases by having the 2nd person select only "good" snakes from those remaining in the basket. The sum of the four probabilities wound up being equal to one of the multiple-choice answers.

Oh, I finally understood how to solve this using the four probabilities and got b) .727 (.7265) as my answer.

Thank you for the responses.