Captain Emily has a ship, the H. M.S Crimson Lynx. The ship is five furlongs from the dread pirate Umaima and her merciless band of thieves. If her ship hasn't already been hit, Captain Emily has probability \dfrac{3}{5} 5 3 start fraction, 3, divided by, 5, end fraction of hitting the pirate ship. If her ship has been hit, Captain Emily will always miss. If her ship hasn't already been hit, dread pirate Umaima has probability \dfrac{1}{7} 7 1 start fraction, 1, divided by, 7, end fraction of hitting the Captain's ship. If her ship has been hit, dread pirate Umaima will always miss. If the Captain and the pirate each shoot once, and the pirate shoots first, what is the probability that the pirate misses the Captain's ship, but the Captain hits?

The probability that the pirate misses the captain's ship but the captain hits = 0.514

Step-by-step explanation:

Let A be the event that the captain hits the pirate ship

The probability of the captain hitting the pirate ship, P(A) = 3/5

Let B be the event that the pirate hits the captain's ship

The probability of the pirate hitting the captain's ship P(B) = 1/7

The probability of the pirate missing the captain's ship, P'(B) = 1 - P(B)

P'(B) = 1 - 1/7 = 6/7

The probability that the pirate misses the captain's ship but the captain hits = P(A) * P(B) = 3/5 * 6/7

= 0.514

5x+10 is the correct one

hope this

take the logarithm of both sides (the base of the log doesn't matter):

apply the exponent property - this says that :

distribute the log terms on both sides:

group like terms together:

divide through both sides by the coefficient on :

you can do some additional rewriting here using the properties of the logarithm to simplify things if you like: