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Answer:
a) P ( E | F ) = 0.54545
b) P ( E | F' ) = 0
Step-by-step explanation:
Given:
- 4 Coins are tossed
- Event E exactly 2 coins shows tail
- Event F at-least two coins show tail
Find:
- Find P ( E | F )
- Find P ( E | F prime )
Solution:
- The probability of head H and tail T = 0.5, and all events are independent
So,
P ( Exactly 2 T ) = ( TTHH ) + ( THHT ) + ( THTH ) + ( HTTH ) + ( HHTT) + ( HTHT) = 6*(1/2)^4 = 0.375
P ( At-least 2 T ) = P ( Exactly 2 T ) + P ( Exactly 3 T ) + P ( Exactly 4 T) = 0.375 + ( HTTT) + (THTT) + (TTHT) + (TTTH) + ( TTTT)
= 0.375 + 5*(1/2)^4 = 0.375 + 0.3125 = 0.6875
- The probabilities for each events are:
P ( E ) = 0.375
P ( F ) = 0.6875
- The Probability to get exactly two tails given that at-least 2 tails were achieved:
P ( E | F ) = P ( E & F ) / P ( F )
P ( E | F ) = 0.375 / 0.6875
P ( E | F ) = 0.54545
- The Probability to get exactly two tails given that less than 2 tails were achieved:
P ( E | F' ) = P ( E & F' ) / P ( F )
P ( E | F' ) = 0 / 0.6875
P ( E | F' ) = 0
Just pay attention to the amount of stripes
The angles with one like equals the one with one, the angle with 2 equals , and so does the one with three
so, ABC will be equal to : MNP
hope this helps
Answer:
a) Mean = 2.6
B) Variance = 9.2
Step-by-step explanation:
We are given the following probability mass function in the question:
a) Mean of random variable
b) Variance of the random variable
