Welcome to our community

Be a part of something great, join today!

Find the probability that in a random sample of 4 shots, he will miss the bullseye at least 3 times

SteffiB

New member
Feb 3, 2019
1
1.Mary has two friends, Ann and Sarah. Mary will visit Ann this evening if the 19A bus arrives before 6PM. Otherwise, She will visit Sarah. The probability of the 19A bus arriving before 6pm is 40%. If she visits Ann, the probability that Ann will be at home is 10% and if she visits Sarah, the probability that Sarah will be home is 20%. Find the probability that the friend Mary visits will be home

2.A darts player finds that on average he hits the bullseye 4 times out of 5. Calculate the probability that in a random sample of 4 shots, he will miss the bullseye at least 3 times
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
464
1.Mary has two friends, Ann and Sarah. Mary will visit Ann this evening if the 19A bus arrives before 6PM. Otherwise, She will visit Sarah. The probability of the 19A bus arriving before 6pm is 40%. If she visits Ann, the probability that Ann will be at home is 10% and if she visits Sarah, the probability that Sarah will be home is 20%. Find the probability that the friend Mary visits will be home
Imagine this happenng 1000 times. 40% of the time, 400 times, she visits Ann. 10% of those, 40 times, Ann is home. 100- 40= 60% of the time, 600 times, she visits Sarah. 20% of those, 120 times, Sarah is home. So of the 1000 times, the person she visits is home 40+ 120= 160 times. The probably the person she visits will be home is 160/1000= 16%.

2.A darts player finds that on average he hits the bullseye 4 times out of 5. Calculate the probability that in a random sample of 4 shots, he will miss the bullseye at least 3 times
Since, on average, he hits the bullseye "4 times out of 5" the probability he will hit the bullseye on any one shot is 4/5 and the probability he misses is 1- 4/5= 1/5. "At least 3 times" in 4 shots is either "all 4 shots" or "3 out 4 shots". The probability of missing "all 4 shots" is (1/5)(1/5)(1/5)(1/5)= 1/5^4= 1/625. One way of missing "3 out of 4 shots" is to miss the first three shots and hit the last one. The probability of that is (1/5)(1/5)(1/5)(4/5)= 4/5^4= 4/635. But you could also miss the first two, hit the third and then miss the fourth. That is also (1/5)(1/5)(4/5)(1/5) which is also 4/625. There are also "Hit, Miss, Hit, Hit" and "Miss, Hit, Hit, Hit". There are four such orders and tne probability of each is 4/625. That means that the probability of "three misses and one hit", in any order, is 4(4/625)= 16/625.

The probability of "at least three misses" is the sum, 1/625+ 16/625= 17/625.