can u do this question on probability?

Probability that Prof A gets heads= 1/2

Probability that Prof B gets heads= 3(1/2*1/2)= 3/4

Probability that Prof B doesn't get heads= 1-3/4 = 1/4

( If Prof A has to get more heads than Prof B, Prof B has to get two tails)

Thus, Probability that Prof A gets more heads than Prof B= 1/2*1/4 = 1/8

1/2 from provability of Prof A getting heads and 1/4 from Prof B getting no heads.

Hope that helped! :)

professor A has a 50% chance of hitting heads and prof. B has a 75% chance

if professor A needs to have more heads than professor B it means that prof. A needs to get heads on his first cointoss and prof. B needs to get tales on both of his.

0.25x0.50=12,5%

where .25 is the chance that prof. B gets two tales and .50 is the chance that prof A gets heads once.

The combinations are,

H-HH

H-HT

H-TH

H-TT.......*

T-HH

T-HT

T-TH

T-TT

and these are all mutually independent, equally possible outcomes . therefore,those marked with * are the ones when prof.A gets more heads than prof. B

so, out of 8 combinitions, 1 satisfy so the probability is 1/8 = 0.125