P(J / R) = P (J and R) / P(R)
<span>0.8 = P (J and R) / 0.6 </span>
<span>P (J and R) = 0.6 * 0.8 = 0.48 [Probability John practicing and it is raining] </span>
<span>P(J / NR) = P (J and NR) / P(NR) </span>
<span>0.4 = P (J and NR) / (1 - 0.6) = P (J and NR) / 0.4 </span>
<span>P (J and NR) = 0.4 * 0.4 = 0.16 [Probability John practicing and it is not raining] </span>
<span>Hence; </span>
<span>Propability of John practicing regardless of weather condition is </span>
<span>P(John Practicing) = 0.48 + 0.16 = 0.64</span>
Answers:
a) 0.0625
b) 0.9375
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Work Shown:
The probability of landing on heads is 1/2 = 0.5 since both sides are equally likely to land on. Getting 4 heads in a row is (1/2)^4 = (0.5)^4 = 0.0625
The event of getting at least one tail is the complement of getting all four heads. This is because you either get all four heads or you get at least one tail. One or the other must happen. We subtract the result we got from 1 to get 1-0.0625 = 0.9375
You can think of it like this
P(getting all four heads) + P(getting at least one tail) = 1
The phrasing "at least one tail" means "one tail or more".
Answer:
only 0
Step-by-step explanation:
8g < 4 simplifies to g < 4/8, i.e., g < 1/2
only 0 is smaller than 1/2
Step-by-step explanation:
The attachment above is a table summarizing everything.
so the answer should be - 0.5
The answer to your questions are......
1.
B. 8........(8 + 7)(8 - 8) = 0
C. -7........(-7 + 7)(-7 - 8) = 0
2.
A. 1/3.......(3*1/3 - 1)(5*1/3 + 20) = 0
C. -4.......(3*-4 - 1)(5*-4 + 20) = 0
Hope this helps! :)
