Al tosses one quarter and at the same time Betty tosses two quarters. What is the probability that Betty gets the same number of heads as Al?
This is what I have:
Al H/T ½ 50%
Betty HH / HT / TH/ TT ½ 50%
The probability is ½ or 50% chance that Betty will get the same amount of heads as Al
Can you see if I'm on the right track or can you guide me to the right track
If I am understanding right both of the probabilities
1/2 * 1/2 = 1/4?
Since there is a 50/50% chance Al will get a head, and there is also a 50/50% chance Betty will get heads on at least on coin. So what I do is what is above, correct?
Schuyla
Will if the probability of Al getting heads is 1/2 and the probability for Betty getting one head is 1/4. The probability of both of them getting 1 head is 1/8.
Okay I'm second guessing myself but with the previous answer I gave you of no heads being 1/8, would that make the possibility 7/8 for a head?
If the probability for Betty getting exactly one head is 1/2 is because she has two quarters and there is a 50% chance that she would get at least one head.
Betty possible outcomes are {HH, HT, TH, TT} 2 out of the 4 tosses have 1 head which equals 1/2