Probability

P(the emergency room is crowded on alternate days in the week)
=
P(the emergency room is crowded on 3 days given at least 2 days)
+ P(the emergency room is crowded on 4 days given at least 2 days)
+ P(the emergency room is crowded on 5 days given at least 2 days)
=
( P(the emergency room is crowded on 3 days)
+ P(the emergency room is crowded on 4 days)
+ P(the emergency room is crowded on 5 days) ) / P(the emergency room is crowded on at least 2 days)
=
( (5C3) (0.3997)³ (1 - 0.3997)² + (5C4) (0.3997)⁴(1 - 0.3997) + (0.3997)⁵ ) / ( 1 - P(the emergency room is crowded on 0 day or 1 day) )
=
( (5C3) (0.3997)³ (1 - 0.3997)² + (5C4) (0.3997)⁴(1 - 0.3997) + (0.3997)⁵ ) / ( 1 - (1 - 0.3997)⁵ - (5C1) (0.3997) (1 - 0.3997)⁴)
= 0.3169217... / 0.6625124...
= 0.478 (3 sig. fig.)

2012-01-29 15:55:22 補充：
Corr :

P(the emergency room is crowded on alternate days in the week)

=
( P(is crowded on day 1 , 3 , 5) + P(is crowded on day 2 , 4) ) / P(is crowded on at least 2 days)

= ( (0.3997)³ (1 - 0.3997)² + (0.3997)² (1 - 0.3997)³ ) / ( 1 - P(the emergency room is crowded on 0 day or 1 day) )

2012-01-29 15:55:28 補充：
= ( (0.3997)³ (1 - 0.3997)² + (0.3997)² (1 - 0.3997)³ ) / ( 1 - (1 - 0.3997)⁵ - (5C1) (0.3997) (1 - 0.3997)⁴)

= 0.05757116... / 0.6625124...

= 0.0869 (3 sig. fig.)

Sorry for my mistakes.

2012-01-29 15:55:53 補充：
Corr :

P(the emergency room is crowded on alternate days in the week)

=
( P(is crowded on day 1 , 3 , 5) + P(is crowded on day 2 , 4) ) / P(is crowded on at least 2 days)

= ( (0.3997)³ (1 - 0.3997)² + (0.3997)² (1 - 0.3997)³ ) / ( 1 - P(the emergency room is crowded on 0 day or 1 day) )

2012-01-29 15:56:05 補充：
= ( (0.3997)³ (1 - 0.3997)² + (0.3997)² (1 - 0.3997)³ ) / ( 1 - (1 - 0.3997)⁵ - (5C1) (0.3997) (1 - 0.3997)⁴)

= 0.05757116... / 0.6625124...

= 0.0869 (3 sig. fig.)

Sorry for my mistakes.