Yes you have to make the assumption that the drops are independent. If you do, you can calculate it like this:
Simplified scenario of 10 drops where on the 3 drops both phones had the same amount of damage.
The binomial probability formula is:
P(X=k)=(nk)⋅pk⋅(1−p)n−kP(X=k)=(kn)⋅pk⋅(1−p)n−k
Where:
- nn is the number of trials (in this case, 10 drops)
- kk is the number of successes (in this case, 7 drops where phone B had more damage)
- pp is the probability of success on any given trial
- (nk)(kn) is "n choose k," a binomial coefficient that can be calculated as n!k!⋅(n−k)!k!⋅(n−k)!n!
For the odds of this occurring by chance, we would assume that there is an equal probability that either phone could have more damage on any given drop, meaning p=0.5p=0.5.
Let's plug in the values and calculate the probability:
- n=10n=10 drops
- k=7k=7 drops where phone B had more damage
- p=0.5p=0.5
P(X=7)=(107)⋅(0.5)7⋅(0.5)3P(X=7)=(710)⋅(0.5)7⋅(0.5)3
P(X=7)=10!7!⋅3!⋅0.510P(X=7)=7!⋅3!10!⋅0.510
P(X=7)=10⋅9⋅83⋅2⋅1⋅0.510P(X=7)=3⋅2⋅110⋅9⋅8⋅0.510 P(X=7)=1206⋅0.510P(X=7)=6120⋅0.510 P(X=7)=20⋅0.0009765625P(X=7)=20⋅0.0009765625 P(X=7)≈0.01953125P(X=7)≈0.01953125
However, to find the probability that phone B has more damage in 7 or more drops (i.e. 7, 8, 9, or 10 drops), we need to find the probability for each of these cases and then sum them up.
So, we also need to calculate:
P(X=8)=(108)⋅(0.5)8⋅(0.5)2P(X=8)=(810)⋅(0.5)8⋅(0.5)2 P(X=9)=(109)⋅(0.5)9⋅(0.5)1P(X=9)=(910)⋅(0.5)9⋅(0.5)1 P(X=10)=(1010)⋅(0.5)10⋅(0.5)0P(X=10)=(1010)⋅(0.5)10⋅(0.5)0
Let's calculate each of these and then sum them all up:
P(X≥7)=P(X=7)+P(X=8)+P(X=9)+P(X=10)P(X≥7)=P(X=7)+P(X=8)+P(X=9)+P(X=10)
P(X≥7)≈0.01953125+(108)⋅0.510+(109)⋅0.510+(1010)⋅0.510P(X≥7)≈0.01953125+(810)⋅0.510+(910)⋅0.510+(1010)⋅0.510
Let's calculate the individual terms:
- P(X=8)≈451024≈0.0439453125P(X=8)≈102445≈0.0439453125
- P(X=9)≈101024≈0.009765625P(X=9)≈102410≈0.009765625
- P(X=10)≈11024≈0.0009765625P(X=10)≈10241≈0.0009765625
P(X≥7)≈0.01953125+0.0439453125+0.009765625+0.0009765625P(X≥7)≈0.01953125+0.0439453125+0.009765625+0.0009765625 P(X≥7)≈0.07421875P(X≥7)≈0.07421875
So, the probability that phone B has more damage in 7 or more drops by chance is approximately 0.07420.0742 or 7.42%7.42%.
Odds are another way of expressing probability and are usually given as a ratio of favorable to unfavorable outcomes. The odds of phone B having more damage in 7 or more drops by chance are:
Odds(X≥7)=P(X≥7)1−P(X≥7)Odds(X≥7)=1−P(X≥7)P(X≥7) Odds(X≥7)≈0.074218751−0.07421875Odds(X≥7)≈1−0.074218750.07421875 Odds(X≥7)≈0.074218750.92578125Odds(X≥7)≈0.925781250.07421875 Odds(X≥7)≈0.0801Odds(X≥7)≈0.0801
This means the odds are approximately 0.0801:1 or about 1:12.5 against this happening by chance.
