Answer:
The chance that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses is P=0.0488.
Step-by-step explanation:
To solve this problem we divide the tossing in two: the first 5 tosses and the last 5 tosses.
Both heads and tails have an individual probability p=0.5.
Then, both group of five tosses have the same binomial distribution: n=5, p=0.5.
The probability that k heads are in the sample is:
Then, the probability that exactly 2 heads are among the first five tosses can be calculated as:
For the last five tosses, the probability that are exactly 4 heads is:
Then, the probability that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses can be calculated multypling the probabilities of these two independent events:
There really isn’t a possible answer since there is not enough information given. Not from what I have learned so far at least, but using common sense, and the process of elimination I would say 30
Answer:IS B OR D BUT I SAY D
Step-by-step explanation:
The answer is 540 because 32/48 is 2/3 and 360/540 equals 2/3