Question 2.6. (DO NOT GET ADDICTED TO VIDEO GAMES) An engineer designed a video

Question

Question 2.6. (DO NOT GET ADDICTED TO VIDEO GAMES) An engineer designed a video game for kids. Inside the program he used five distinct birds, namely: Bluebird, Cardinals, Dove, Eagle, and Flamingo. The engineer allows only three birds to appear on the screen in a row and selection is made on purely random basis using a random number table. Flamingo Cardinal Dove Eagle Fig, 2.6. Birds in the game Bluebird Amy plays the game on a computer, and she wonders: (a) What is the probability that same Bluebird will appear three times on the screen? Fig. 2.7. Amy (b) What is the probability that same bird will appear three times? ) Assume the engineer did not allow for the same bird to appear three times, what is the probability that Bluebird, Dove and Flamingo will appear on the screen at the first, second and third places?

Explanation / Answer

Question 2.6

There are 5 different birds and only 3 birds in a row can appear.

(a) Three bluebirds will appear on the screen

Total number of possibilities for birds to appear on screen = 53 = 125 [ as at each place in row there are 5 possibilities that a bird can appear]

so Probability that same bluebird will appear on screen 3 times = 1/125

(b) ANy bird will appear three times on screen

Total number of possibilities for birds to appear on screen = 53 = 125

Possibilities that any bird will appear on screen three times = 5

so Probability of any bird will appear on screen three times = 5/125 = 1/25 = 0.04

(c) Engineer don't allow same bird to appear three times. So, number of such possibilities = 125 - 5 = 120

Now all three places have been decided so probability of that thing to happen = 1/120

(d) Here the order is not important so they can be anywhere. Now one of these three has occupied a place that wiill leave only 2 places for the second bird and only 1 place for 3rd bird.

So total number of combinations these three birds will catch = 3 * 2 * 1 = 6

so Probability of such thing to occur = 6/120 = 1/20

(e) Now now bird can appear more than once so total number of possibilities are = 5P3 = 5 * 4 * 3 = 60

now the probabilty that these three chosen birds will get the place as decided = 1/60

(f) Here the order is not important so they can be anywhere. Now one of these three has occupied a place that wiill leave only 2 places for the second bird and only 1 place for 3rd bird.

So total number of combinations these three birds will catch = 3 * 2 * 1 = 6

so Probability of such thing to occur = 6/60 = 1/10

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