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
Answer:
you can solve problems by this equation
Hope this makes sense! It was too hard to type out lol. The one on the left is how it is solved the and one on the right kinda explains how I got the equation.
Answer: D. 20/3
Step-by-step explanation:
20/3 = 6.6 (repeating)
