B. Every third student entering the school in the morning
3/5 squared in 9/25, 3*3/5*5 :)
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
-3(x-1)/x would be your answer
From the given mean and margin of error, the 99% confidence interval for the mean amount of money spent on lunch per week for all students is:
[$19.5, $22.5].
<h3>How to calculate a confidence interval given the sample mean and the margin of error?</h3>
The confidence interval is given by the sample mean plus/minus the margin of error, hence:
- The lower bound is the sample mean subtracted by the margin of error.
- The upper bound is the sample mean added to the margin of error.
For this problem, we have that:
- The sample mean is of $21.
- The margin of error is of $1.50.
Hence the bounds are given as follows:
- Lower bound: 21 - 1.50 = $19.50.
- Upper bound: 21 + 1.50 = $22.50.
Hence the interval is [$19.50, $22.50].
More can be learned about confidence intervals at brainly.com/question/25890103
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