A player of a video game isconfronted with a series of 3 opponents and a 70% pro

Question

A player of a video game isconfronted with a series of 3 opponents and a 70% probabilityof defeating each opponent. Assume that the results from opponentsare independent( and that when the player is defeated by anopponent the game ends). a) what is the probability that a player defeats at least twoopponents in a game? b)If the game is played 5 times, what is the probability thatthe player defeats all 3 opponents at least twice? A player of a video game isconfronted with a series of 3 opponents and a 70% probabilityof defeating each opponent. Assume that the results from opponentsare independent( and that when the player is defeated by anopponent the game ends). a) what is the probability that a player defeats at least twoopponents in a game? b)If the game is played 5 times, what is the probability thatthe player defeats all 3 opponents at least twice?

Explanation / Answer

probability of defeating an opponent = 0.7 probability of not defeating an opponent = 0.3 a) probability of defeating atleast 2 opponents = probabilityof defeating exactly 2 opponents + prob. of defeating 3opponents probability of defeating 2 opponents (i.e the opponents whomthe player defeats can be the 1st, 2nd opponent, 2nd, 3rd opponentor 3rd, 1st opponent, i.e there are 3 possible combinations ofopponents) so probability of the above = 3 x 0.7 x 0.7 x 0.3 =0.441 probability of defeating all 3 = 0.7 x 0.7 x 0.7 = 0.343 hence probability of (a) = 0.441 +0.343 =0.784 b) probability of defeating all three = 0.343 hence probability of not defeating all three =1-0.343=0.657 in 5 games probability that 3 opponents are defeated 0 times =0.657 x 0.657 x 0.657 x 0.657 x 0.657 =0.1224 in 5 games probability that 3 opponents are defeated only 1time = 5 x 0.343 x 0.657 x0.657x0.657x0.657 =0.32 hence probability that if a game is played 5 times, the playerdefeats all 3 opponents atleast twice = 1-(0.1224 +0.32) = 0.5576

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