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 :
![P(X=r)=^nC_r\left(\dfrac{1}{2}\right)^r\left(\dfrac{1}{2}\right)^{n-r}.](https://tex.z-dn.net/?f=P%28X%3Dr%29%3D%5EnC_r%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5Er%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7Bn-r%7D.)
(a) The probability of getting 5 heads is given by
![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.](https://tex.z-dn.net/?f=P%28X%3D5%29%5C%5C%5C%5C%5C%5C%3D%5E%7B10%7DC_5%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E5%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B10-5%7D%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7B10%21%7D%7B5%21%2810-5%29%21%7D%5Cdfrac%7B1%7D%7B2%5E%7B10%7D%7D%5C%5C%5C%5C%5C%5C%3D0.24609%5C%5C%5C%5C%5Csim0.25.)
(b) The probability of getting 5 or more than 5 heads is
![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.](https://tex.z-dn.net/?f=P%28X%5Cgeq%205%29%5C%5C%5C%5C%3DP%28X%3D5%29%2BP%28X%3D6%29%2BP%28X%3D7%29%2BP%28X%3D8%29%2BP%28X%3D9%29%2BP%28X%3D10%29%5C%5C%5C%5C%3D%5E%7B10%7DC_5%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E5%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B10-5%7D%2B%5E%7B10%7DC_6%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E6%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B10-6%7D%2B%5E%7B10%7DC_7%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E7%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B10-7%7D%2B%5E%7B10%7DC_8%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E8%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B10-8%7D%2B%5E%7B10%7DC_9%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E9%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B10-9%7D%2B%5E%7B10%7DC_%7B10%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B10%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B10-10%7D%5C%5C%5C%5C%5C%5C%3D0.24609%2B0.20507%2B0.11718%2B0.04394%2B0.0097%2B0.00097%5C%5C%5C%5C%3D0.62295%5C%5C%5C%5C%5Csim%200.62.)
(c) Proceeding as in parts (a) and (b), we see that
if a = 10, then
![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.](https://tex.z-dn.net/?f=P%28X%5Cleq%200%29%3D0.00097%2C%5C%5C%5C%5CP%28X%5Cleq%201%29%3D0.01067%2C%5C%5C%5C%5CP%28X%5Cleq%202%29%3D0.05461%2C%5C%5C%5C%5CP%28X%5Cleq%203%29%3D0.17179%2C%5C%5C%5C%5CP%28X%5Cleq%204%29%3D0.37686%2C%5C%5C%5C%5CP%28X%5Cleq%205%29%3D0.62295%2C%5C%5C%5C%5CP%28X%5Cleq%206%29%3D0.82802.)
Therefore, the minimum value of a is 6.
Hence, all the questions are answered.