Two baseball players bat first and second in the lineup. The first batter has an on-base percentage of 0.23. The second batter has an on-percentage of 0.38 if someone is on base, but only 0.26 if the bases are empty. At the start of the game, what is the probability that neither player gets on base?

The probability that neither player gets on base is 0.4824

Step-by-step explanation:

1. Both players get to base. Just multiply the two probabilities together:

= (probability first batter gets on base) x (probability second batter gets on base, if the first batter gets on base)

= 0.23 x 0.38

= 0.0874

2. One player gets to base. The formula here is P(A+B) =P(A) + P(B) - P(A) x P(B)

= (probability first batter gets on base) + (probability second batter gets on base, if the first batter does not) - (0.23 x 0.26)

= 0.23 + 0.26 - (0.23 x 0.26)

= 0.49 - 0.0598

= 0.4302

3. Neither player gets to base = 1 - addition of the previous two cases.

= 1 - (0.0874 + 0.4302)

= 1 - 0.5176

= 0.4824

the answer is mp

step-by-step explanation: