Lets apply basic properties of probability on a batter. What would be the
probability of a three-hundred batter making a hit first at bat, getting out
in second at bat, making another hit in third at bat?
If event A is making a hit,P(A)= 130 , P(AC)= 170. Since the events are
independent, P(A)×P(AC)×P(A)= 130 × 170 × 130 = 160300.
What about the probability of a batter with 0.333 average making a
single hit in three times at bat?
If event H is making a hit, event O is getting out,
P(H)= 1 / 3 , P(HC)=P(O)= 2 / 3. There are three combination for a
batter to hit one hit in three times at bat which are HOO, OHO, OOH
( 3 C 1 =^31 P!^1 = 3 ) and probability of each combination are
31 × 32 × 32 = 247 .So total probability is 247 + 247 + 247 = 94.^
Lets calculate the probability of the same batter hitting 4 hits in 10 times
at bat.
If event H is making a hit, event O is getting out,
P(H)= 1 / 3 , P(O)= 2 / 3. Combination of hitting 4 hits in 10 at bats is
10 C 4. Probability of each combination are (^13 )^4 ( 32 )^6 Therefore,
probability of hitting 4 hits in 10 times at bat are
10 C 4 ×( 31 )^4 ×(^23 )^6 =^13341400 = 5139044490 = 149468803. This form of calculations
is called probability of independent experiments, also known as
Bernoulli’s trial.
If a probability of event A is P in an experiment, in an independent