Question

a biased coin is tossed 6 times the probability of heads on any toss is 0.3. let x denote the numbers of head that comes up. Find 1.p(x=2) 2.p(x>2) 3.p(0

Answer 1

Answer:

sorry I didn't understand your question

Answer 2

Answer: P(x=2) =0.3241

2.P(x>2)= 0.2557

3.p(0)=0.11649

Step-by-step explanation:

• The Concept used in the question is binomial theorem.

• The theorem is used for expanding of the exponential variables.

• The formula of binomial theorem is

        $P=^nC_{r} p^rq^n^-^r$

       where n is number of trials, r is the number of success, p is

       probability of success and q is the probability of failure

Step 1 of 1:

• Probability of success=0.3
• Probability of failure=0.7

1. Probability when r=2

         $P=^nC_{r} p^rq^n^-^r$

        $P=^6C_{2}\ 0.3^2\ \ 0.7^6^-^2$

        $P=\frac{6\times5}{2} \times0.09\times0.2401\\P=0.3241$

      2.   Probability when X=0

         $P=^nC_{r} p^rq^n^-^r$

        $P=^6C_{0}\ 0.3^0\ \ 0.7^6^-^0$

        $P=1\times1\times0.117649\\P=0.117649$

       3. Probability when X>2=1-P(0)-P(1)-P(2)

         $P=^nC_{r} p^rq^n^-^r$

        $P=1-^6C_{0}\ 0.3^0\ \ 0.7^6^-^0\ \ -^6C_{1}\ 0.3^1\ \ 0.7^6^-^1-^6C_{2}\ 0.3^2\ \ 0.7^6^-^2\\P=1-0.1176-.302526-.324125\\P=0.2557$