Suppose that in an alternate universe, the gambler's fallacy is true: the more a gambler loses, the more likely she is to win the next game. To be precise, if the gambler loses a game, she will win the next game with probability 3/4. If the gambler wins the game, she will win the next game with probability 1/2.

Answer and explanation:

The gambler's fallacy is the fallacy of belief that if an event such as a loss occurs more frequently in the past, it is less likely to happen in the future. We assume here that this belief is true, therefore

If she loses, her probability of winning increases =3/4

If she wins, her probability to win is normal =1/2

Given that probability of winning is 1/2

Probability of losing is 1-1/2=1/2

Probability that she wins the tournament is probability that she wins the first two games and loses the last or wins the first game, loses the second and wins the last or loses the first game and wins the last two games or probability that she wins all three games

=1/2*1/2*1/2+1/2*1/2*3/4+1/2*3/4*1/2+1/2*1/2*1/2

=25/48

Probability of winning the tournament if she loses the first game

=1/2*3/4*1/2= 3/16

Note: whenever there is "or" in probability, you add

l.jnl.j/n.m,lm,

step-by-step explanation:

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm