Question

find the probability given the situation. (see example 2)

1. a game show airs on television five days per week, with a new player each day. each day, a prize is randomly placed behind one of two doors. the contestant wins the prize by selecting the correct door. what is the probability that exactly two of the five contestants win a prize during a week? hint: use the possible outcomes chart below.

possible outcomes

number of winners	outcomes
1	wllll lwlll llwll lllwl llllw
2	wwlll wlwl l wllwl wllwl lwwll lwlwl lwllw llwwl llwlw lllww
3	wwwll wwlwl wwllw wlwwl wlwlw wllww lwwwl lwwlw lwlww llwww
4	wwwwl wwwlw wwlww wlwww lwwww
5	wwwww

Explanation:

Step1: Count total outcomes

Each day has 2 outcomes (W or L), so for 5 days, total outcomes = \(2^5 = 32\).

Step2: Count outcomes with 2 winners

From the table, for 2 winners:

• First row: 3 outcomes (WWLLL, WLWL L, WLLWL)
• Second row: 2 outcomes (WLLLW, LWWLL)
• Third row: 2 outcomes (LWLWL, LWLLW)
• Fourth row: 3 outcomes (LLWWL, LLWLW, LLLWW)

Wait, let's count correctly: WWLLL, WLWL L (wait, typo? WLWWL? No, original table: WWLLL, WLWL L (maybe WLWWL? No, the table has WWLLL, WLWL L (wait, the user's table: for 2, first line WWLLL, WLWL L, WLLWL; second line WLLLW, LWWLL, LWLWL; third line LWLLW, LLWWL, LLWLW, LLLWW. Wait, let's count each:

First line: 3 (WWLLL, WLWL L (maybe WLWWL? No, original: WWLLL, WLWL L, WLLWL) – no, maybe the user's table has:

WWLLL, WLWL L (wait, maybe WLWWL? No, the correct count: let's list all for 2 winners:

WWLLL, WLWL L (no, the table as given:

"WWLLL WLWL L WLLWL

WLLLW LWWLL LWLWL

LWLLW LLWWL LLWLW LLLWW"

Wait, first line: 3, second line: 3, third line: 4? Wait no, let's count:

First line: WWLLL, WLWL L, WLLWL → 3

Second line: WLLLW, LWWLL, LWLWL → 3

Third line: LWLLW, LLWWL, LLWLW, LLLWW → 4

Wait, no, that can't be. Wait, the number of combinations for 2 wins in 5 days is \( \binom{5}{2} = 10 \)? Wait no, \( \binom{5}{2} = 10 \). Wait the table: let's count the outcomes for 2 winners:

Looking at the table:

For 2 winners:

First group: WWLLL, WLWL L, WLLWL → 3

Second group: WLLLW, LWWLL, LWLWL → 3

Third group: LWLLW, LLWWL, LLWLW, LLLWW → 4

Wait 3+3+4=10? Wait 3+3=6, +4=10. Yes, \( \binom{5}{2} = 10 \). So total outcomes with 2 winners is 10.

Step3: Calculate probability

Probability = (Number of favorable outcomes) / (Total outcomes) = \( \frac{10}{32} = \frac{5}{16} \). Wait, wait, total outcomes: each day 2 choices, 5 days: \(2^5 = 32\). Number of favorable (exactly 2 wins) is \( \binom{5}{2} = 10 \). So probability is \( \frac{10}{32} = \frac{5}{16} \).

Wait, let's check the table:

For 0 winners: 1 outcome (LLLLL)

1 winner: 5 outcomes (as listed: WLLLL, LWLLL, LLWLL, LLLWL, LLLLW) → 5, which is \( \binom{5}{1} = 5 \)

2 winners: let's count the table's entries:

First line: 3, second line: 2, third line: 5? No, the table as given:

"WWLLL WLWL L WLLWL

WLLLW LWWLL LWLWL

LWLLW LLWWL LLWLW LLLWW"

Wait, first line: 3, second line: 3, third line: 4 → 3+3+4=10, which matches \( \binom{5}{2}=10 \)

3 winners: let's check, \( \binom{5}{3}=10 \), table has:

"WWWLL WWLWL WWLLW

WLWWL WLWLW WLLWW

LWWWL LWWLW LWLWW LLWWW"

First line: 3, second line: 3, third line: 4 → 3+3+4=10, correct.

4 winners: \( \binom{5}{4}=5 \), table has 5 outcomes (as listed: WWWWL, WWWLW, WWLWW, WLWWW, LWWWW) → 5, correct.

5 winners: 1 outcome (WWWWW) → correct.

So total outcomes: 1 (0) +5 (1)+10 (2)+10 (3)+5 (4)+1 (5) = 32, which is \(2^5=32\), correct.

So number of outcomes with exactly 2 winners is 10.

Thus, probability = 10/32 = 5/16.

Answer:

\(\frac{5}{16}\)