Step-by-step explanation:
at least 2 out of the 4 attempts land on heads.
a probability is always the ratio
desired cases / totally possible cases.
for 4 such attempts we have
2⁴ = 16
different possible outcomes in total. as each attempt has 2 possible outcomes, we have 2×2×2×2 = 2⁴ = 16.
so, now our desired outcomes :
head head head head
head head head tail
head head tail head
head tail head head
tail head head head
head head tail tail
head tail head tail
head tail tail head
tail head head tail
tail head tail head
tail tail head head
11 outcomes
in mathematical terms
combinations (as the sequence of the individual tails does not matter, only that there are that many tails) of 4 out of 4, 3 out of 4 and 2 out of 4 :
4!/(4! × (4-4)!) = 4!/4! = 1
4!/(3! × (4-3)!) = 4!/3! = 4
4!/(2! × (4-2)!) = 4!/(2! × 2!) = 4!/(2×2) = 4×3×2/4 = 6
6 + 4 + 1 = 11 outcomes
so, the probability of getting at lest 2 heads in 4 attempts is
11/16 = 0.6875
Answer:
[tex]\dfrac{11}{16}=0.6875=68.75\%[/tex]
Step-by-step explanation:
To find the probability that at least two out of four coin flips land on heads, we first determine the possible outcomes and then count the favorable ones.
As each coin flip has 2 possible outcomes (heads or tails), for 4 coin flips, the total number of possible outcomes is:
[tex]\textsf{Total number of possible outcomes}= 2^4\\\\\textsf{Total number of possible outcomes}= 16[/tex]
To count the number of outcomes with at least 2 heads, we need to determine the number of ways to choose exactly 2, 3, and 4 heads.
To do this, we can use the binomial coefficient, which gives us the number of ways of picking "r" unordered outcomes (heads) from "n" trials (flips):
[tex]\displaystyle \binom{n}{r}= \:^{n}C_{r} = \frac{n!}{r!(n-r)!}[/tex]
For exactly 2 heads:
[tex]\displaystyle \binom{4}{2} = \dfrac{4!}{2!(4-2)!}=\dfrac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}=\dfrac{4\times 3}{2\times 1}=\dfrac{12}{2}=6[/tex]
For exactly 3 heads:
[tex]\displaystyle \binom{4}{3} = \dfrac{4!}{3!(4-3)!}=\dfrac{4\times 3\times 2\times 1}{3\times 2\times 1\times 1}=\dfrac{4}{1}=4[/tex]
For exactly 4 heads:
[tex]\displaystyle \binom{4}{4} = \dfrac{4!}{4!(4-4)!}=\dfrac{4\times 3\times 2\times 1}{4\times 3\times 2\times 1\times 1}=\dfrac{24}{24}=1[/tex]
Therefore, the total number of favorable outcomes is:
[tex]\textsf{Total favorable outcomes}=\displaystyle \binom{4}{2}+ \binom{4}{3}+ \binom{4}{4} \\\\\textsf{Total favorable outcomes}=6+4+1 \\\\ \textsf{Total favorable outcomes}=11[/tex]
To find the probability that at least two out of four coin flips land on heads, P(X ≥ 2), divide the number of favorable outcomes (11) by the total number of possible outcomes (16):
[tex]\rm P(X\geq 2) = \dfrac{11}{16}[/tex]
So, the probability that at least two of the four coin flips will land on heads is:
[tex]\LARGE\boxed{\boxed{\dfrac{11}{16}}}[/tex]
