Ten unbiased coins are tosssimultaneously. Find the probabilityobtaini

1.	Ten unbiased coins are tosssimultaneously. Find the probabilityobtaining.(i) Exactly 6 Heads(ii) Atleast 8 Heads(iii) No Heads(iv) Atleast one Headv) Not more than 3 Heads andvi) Atleast 4 heads.
Answer» 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 * $0.5^{6+<klux>4</klux>}$ = 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 .

Ten unbiased coins are tosssimultaneously. Find the probabilityobtaining.(i) Exactly 6 Heads(ii) Atleast 8 Heads(iii) No Heads(iv) Atleast one Headv) Not more than 3 Heads andvi) Atleast 4 heads.

Answer»

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 * $0.5^{6+<klux>4</klux>}$ = 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 .