Probability and likelihood

a team of scientists is studying the animals at a nature reserve. They capture the animals, mark them so they can identify each animal, and then release them back into the park. The table gives the number of animals they’ve identified. Use this information to complete the two tasks that follow.

animal total in park number marked

elk 5,625 225

wolf 928 232

cougar 865 173

bear 1,940 679

mountain goat 328 164

deer 350 105

moose 215 86

part a

what is the probability of the next elk caught in the park being unmarked? write the probability as a fraction, a decimal number, and a percentage.

font sizes

characters used: 310 / 15000

part b

describe the likelihood of the next elk caught being unmarked.

font sizes

characters used: 58 / 15000

part c

describe a simulation that you can use to model this situation.

font sizes

characters used: 374 / 15000

part d

what is the probability of the next wolf caught in the park being unmarked? write the probability as a fraction, a decimal number, and a percentage.

font sizes

characters used: 54 / 15000

part e

describe the likelihood of the next wolf caught being unmarked.

font sizes

characters used: 0 / 15000

part f

describe a simulation that you can use to model this situation. The simulation should be different from the one in part c.

font sizes

characters used: 86 / 15000

part g

in the unit, you found the probability of a compound event by identifying the sample space. However, it is also possible to find the probability of a compound event without finding the sample space. To do this, multiply the probability of the first event by the probability of the second event. For example, the probability of flipping heads twice on a coin is. Using this idea, what is the probability that the next cougar and bear caught will both be unmarked?

font sizes

characters used: 0 / 15000

part h

describe the likelihood that the next cougar and bear caught are both unmarked.

font sizes

characters used: 0 / 15000

part i

describe a simulation that you can use to model this event.

font sizes

characters used: 0 / 15000

part j

using the method described in part g, what is the probability that the next mountain goat, deer, and moose caught are all unmarked?

font sizes

characters used: 0 / 15000

part k

describe the likelihood that the next mountain goat, deer, and moose caught are all unmarked.

font sizes

characters used: 0 / 15000

part l

describe a simulation that you can use to model this event. Your simulation should be different from the one in part i.

font sizes

characters used: 0 / 15000

The likely hood of the next elk being unmarked is 96/100= 0.96 as 0.96 is close to 1 it is very likely that the next elk will probably be unmarked.

Part C:

A simulation that could model this situation would be:

Take a number from 1 to 100

If it is between 1-96 it would mean that the Elk was unmarked.

If it would be 97-100 it would mean that the Elk was already marked.

Part D:

Fraction: 696/928

Decimal Number: 0.75

Percentage: 75%

Part E:

The likely hood of the next wolf being unmarked is 75/100= 0.75 as 0.75 is close to 1 it is likely that the next wolf will probably be unmarked.

Part F:

You could use a spinner and divide the spinner into 4 colors, Red, Blue, Orange, Yellow. Yellow, Orange, and Red would represent the next wolf being unmarked, and Blue would represent the next wolf being marked.

Part G:

Probability: 52/100 = 0.52 = 52%

Part H:

The likely hood of the next bear and cougar being unmarked is 52/100 = 0.52 as 0.52 is not close nor far away from 1 it is neither likely nor unlikely that the next bear and cougar will be unmarked.

Part I:

You could ask a computer for a number between 1-100

Number 1-52 equals the cougar and bear caught both being unmarked.

Number 53-100 equals either the cougar or bear caught being marked.

Part J:

The probability of the mountain goat, deer, and moose being caught unmarked is 21%

Part K:

The likely hood of the next mountain goat, deer, and moose being unmarked is 21/100 = 0.21 as 0.21 is not close away from 1 it is unlikely that the next mountain goat, deer, and moose will be unmarked.

Part L:

You could use a random number generator to generate numbers between 1-100.

1-21 means that mountain goat, deer, and moose are caught unmarked

22-100 means that mountain goat, deer, and moose are caugth marked.