Answer:
Step-by-step explanation:
Given that,
Outcomes Heads= 4heads and
Outcomes Tails= 6tails
Total outcome of events n=10
Probability of head in is first throw is, the is the outcome of head divide by the total possible outcomes
P(Head in first throw)=4/10
P(H)=2/5
P(H) = 0.4
P(H) =40%
Answer: The minimum reliability for the second stage be 0.979.
Step-by-step explanation:
Since we have given that
Probability for the overall rocket reliable for a successful mission = 97%
Probability for the first stage = 99%
We need to find the minimum reliability for the second stage :
So, it becomes:
P(overall reliability) = P(first stage ) × P(second stage)

Hence, the minimum reliability for the second stage be 0.979.
I think the answer you have there is correct
Set up the equation so m<HCB + m<DCH = m<DCB
2x + 100 + 60 = x + 170
2x+ 160 = x +170
-x -x
x + 160 = 170
- 160 -160
x = 10