Question

Ten unbiased coins are toss
simultaneously. Find the probability
obtaining.
(i) Exactly 6 Heads
(ii) Atleast 8 Heads
(iii) No Heads
(iv) Atleast one Head
v) Not more than 3 Heads and
vi) Atleast 4 heads.​

Answer 1

Answer:

(i) 0.2051

(ii) 0.0547

(iii) 0.0009765

(iv) 0.99902

(v) 0.1720

(vi) 0.828

Step-by-step explanation:

The binomial expression is given by;

$P(X = r) = \binom{n}{r}p^{r}(1-p)^{n-r} for x = 0,1,2,3,.........$

where, Number of trials, n = 10 coins

In all cases we are interested in getting head on a coin so, the probability of success(head), p = 0.5

r = number of successes

(i) Probability of obtaining Exactly 6 Heads = P(X = 6)

    P(X = 6) = $\binom{10}{6}(0.5)^{6}(1-0.5)^{10-6}$ = 210 * $0.5^{6+4}$ = 210 * $0.5^{10}$ = 0.2051

(ii) Probability of obtaining at least 8 Heads = P(X >= 8)

     P(X >=8) = P(X = 8) + P(X = 9) + P(X = 10) =                   $\binom{10}{8}(0.5)^{8}(1-0.5)^{10-8} + \binom{10}{9}(0.5)^{9}(1-0.5)^{10-9}+\binom{10}{10}(0.5)^{10}(1-0.5)^{10-10}$

                   = 45 * $0.5^{10}$ + 10 * $0.5^{10}$ + 1 * $0.5^{10}$ = 0.0547

(iii) Probability of obtaining No Heads = P(X = 0)  

     P(X = 0) = $\binom{10}{0}(0.5)^{0}(1-0.5)^{10-0}$ = 1 * $0.5^{10}$ = 0.0009765

(iv)  Probability of obtaining at least one Head = P(X >= 1)  

      P(X >=1) = 1 - Probability of obtaining No Heads

                    = 1 - P(X = 0) = 1 - 0.0009765 = 0.99902

(v) Probability of obtaining Not more than 3 Heads = P(X <= 3)

    P(X <= 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) =

$\binom{10}{0}(0.5)^{0}(1-0.5)^{10-0}+\binom{10}{1}(0.5)^{1}(1-0.5)^{10-1}+\binom{10}{2}(0.5)^{2}(1-0.5)^{10-2}+\binom{10}{3}(0.5)^{3}(1-0.5)^{10-3}$

   = 1 * $0.5^{10}$ + 10 * $0.5^{10}$ + 45 * $0.5^{10}$ + 120 * $0.5^{10}$ = $0.5^{10}$ * 176 = 0.1720

(vi) Probability of obtaining at least 4 Heads = 1 - Probability of obtaining not more than 3 Heads

     P(X >= 4) = 1 - P(X <= 3) = 1 - 0.1720 = 0.828 .