Question

The probability that Casey, a softball player gets a hit during a single time at bat is 3/10. If she goes to bat 6 times in her next game, determine the probability to four decimal places a. That she gets at least 3 hits in the game. b. That she gets at most 4 hits in the game c. That she does not get a hit in the game. d. That she gets exactly 3 hits in the game

Answer 1

As per the given information;

p = Probability of each success = 3 / 10 = 0.3

q = Probability of each failure = 1 - p = 1 - 0.3 = 0.7

n = Total number of events = 6

r = Number of successful events

Using Binomial Distribution;

P(X=r) = ⁿC_r  p^r q^{(n - r)}

a) At least 3 hits in the game:

Here r  $\geq$ 3

P( r $\geq$ 3) = 1 - [ P(r = 0) + P(r = 1) + P(r = 2) ]

= 1 - [ ⁶C₀(0.3)⁰(0.7)^{(6 - 0)} + ⁶C₁ (0.3)¹(0.7)^{(6 - 1)} +  ⁶C₂ (0.3)²(0.7)^{(6 - 2)} ]

= 1 - [ 0.11765 +  0.3025 + 0.3241 }

= 1 - 0.74425

= 0.25575

b) At most 4 hits in the game:

Here r  $\leq$ 4

P( r  $\leq$ 4) = P(r = 0) + P(r = 1) + P(r = 2) + P( r = 3) + P( r = 4)

=  ⁶C₀ (0.3)⁰(0.7)^{(6 - 0)} + ⁶C₁ (0.3)¹(0.7)^{(6 - 1)} +  ⁶C₂ (0.3)²(0.7)^{(6 - 2)} + ⁶C₃ (0.3)³(0.7)^{(6 - 3)} + ⁶C₄ (0.3)⁴(0.7)^{(6 - 4)}

= 0.11765 +  0.3025 + 0.3241 + 0.1852 + 0.05953

= 0.98898

c) No hits in the game:

Here r = 0

P( r = 0) = ⁶C₀ (0.3)⁰(0.7)^{(6 - 0)}

= 0.11765

d) Exactly 3 hits in the game:

Here r = 3

P( r = 3) = ⁶C₃ (0.3)³(0.7)^{(6 - 3)}

= 0.1852