Question

Ten coins are thrown simultaneously. find the probability of getting at least 6 heads

Answer 1

Probability of getting at least 6 heads is 0.38 .

Step-by-step explanation:

We are given that Ten coins are thrown simultaneously and we have to find the probability of getting at least 6 heads.

For calculating above probability we will consider Binomial distribution;

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

where,  n = number of trials = 10 coins

             r = number of success = at least 6  

             p = probability of success and success in our question is getting a

                    head , i.e.; p = 0.5 {because there is 50-50 chance of getting a

                       head on the coin}

Let X = Number of heads

So,  the probability of getting at least 6 heads is given by = P(X $\geq$ 6)

P(X $\geq$ 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

= $\binom{10}{6} 0.5^{6} (1-0.5)^{10-6}+ \binom{10}{7} 0.5^{7} (1-0.5)^{10-7}+\binom{10}{8} 0.5^{8} (1-0.5)^{10-8}+\binom{10}{9} 0.5^{9} (1-0.5)^{10-9}+\binom{10}{10} 0.5^{10} (1-0.5)^{10-10}$

= $210 \times 0.5^{6} \times 0.5^{4} +120 \times 0.5^{7} \times 0.5^{3} +45 \times 0.5^{8} \times 0.5^{2} +10 \times 0.5^{9} \times 0.5^{1} +1 \times 0.5^{10} \times 0.5^{0}$

= $210 \times 0.5^{10} +120 \times 0.5^{10} +45 \times 0.5^{10} +10 \times 0.5^{10} +1 \times 0.5^{10}$

= $0.5^{10} \times [210 +120 +45 +10 +1 ]$

= $0.5^{10} \times 386$ = 0.38

Therefore, probability of getting at least 6 heads is 0.38.