Respuesta :
Answer: The required answers are
(a) 0.25, (b) 0.62, (c) 6.
Step-by-step explanation: Given that we toss a fair coin 10 times and X denote the number of heads.
We are to find
(a) the probability that X=5
(b) the probability that X greater or equal than 5
(c) the minimum value of a such that P(X ≤ a) > 0.8.
We know that the probability of getting r heads out of n tosses in a toss of coin is given by the formula of binomial distribution as follows :
[tex]P(X=r)=^nC_r\left(\dfrac{1}{2}\right)^r\left(\dfrac{1}{2}\right)^{n-r}.[/tex]
(a) The probability of getting 5 heads is given by
[tex]P(X=5)\\\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}\\\\\\=\dfrac{10!}{5!(10-5)!}\dfrac{1}{2^{10}}\\\\\\=0.24609\\\\\sim0.25.[/tex]
(b) The probability of getting 5 or more than 5 heads is
[tex]P(X\geq 5)\\\\=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}+^{10}C_6\left(\dfrac{1}{2}\right)^6\left(\dfrac{1}{2}\right)^{10-6}+^{10}C_7\left(\dfrac{1}{2}\right)^7\left(\dfrac{1}{2}\right)^{10-7}+^{10}C_8\left(\dfrac{1}{2}\right)^8\left(\dfrac{1}{2}\right)^{10-8}+^{10}C_9\left(\dfrac{1}{2}\right)^9\left(\dfrac{1}{2}\right)^{10-9}+^{10}C_{10}\left(\dfrac{1}{2}\right)^{10}\left(\dfrac{1}{2}\right)^{10-10}\\\\\\=0.24609+0.20507+0.11718+0.04394+0.0097+0.00097\\\\=0.62295\\\\\sim 0.62.[/tex]
(c) Proceeding as in parts (a) and (b), we see that
if a = 10, then
[tex]P(X\leq 0)=0.00097,\\\\P(X\leq 1)=0.01067,\\\\P(X\leq 2)=0.05461,\\\\P(X\leq 3)=0.17179,\\\\P(X\leq 4)=0.37686,\\\\P(X\leq 5)=0.62295,\\\\P(X\leq 6)=0.82802.[/tex]
Therefore, the minimum value of a is 6.
Hence, all the questions are answered.
