Question

thth
thht
tthh
ttht
ttth
tttt
what is the probability of at least two coins landing on heads?
$\frac{5}{16}$
$\frac{3}{8}$
$\frac{1}{2}$
$\frac{11}{16}$

Explanation:

Step1: Find total number of outcomes

Each coin - toss has 2 outcomes. For 4 coin - tosses, the total number of outcomes is $2\times2\times2\times2 = 16$ by the multiplication principle.

Step2: Count number of favorable outcomes

List the number of heads in each outcome:

• THTH: 2 heads
• THTT: 1 head
• TTHH: 2 heads
• TTHT: 1 head
• TTTH: 1 head
• TTTT: 0 heads

The favorable outcomes (at least 2 heads) are THTH, TTHH.
We can also calculate the number of favorable outcomes by considering cases:

• 2 heads: The number of ways to get 2 heads out of 4 coin - tosses is $C(4,2)=\frac{4!}{2!(4 - 2)!}=\frac{4\times3\times2!}{2!\times2!}=6$
• 3 heads: The number of ways to get 3 heads out of 4 coin - tosses is $C(4,3)=\frac{4!}{3!(4 - 3)!}=\frac{4\times3!}{3!\times1!}=4$
• 4 heads: The number of ways to get 4 heads out of 4 coin - tosses is $C(4,4)=\frac{4!}{4!(4 - 4)!}=1$

The total number of favorable outcomes is $C(4,2)+C(4,3)+C(4,4)=6 + 4+1=11$

Step3: Calculate probability

The probability $P$ of an event is given by $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. So $P=\frac{11}{16}$

Answer:

$\frac{11}{16}$