Question

An unbiased coin tossed n times.Probability of getting 4 tails equals the probability of getting 7 tails. Then the probability of getting 2 tails are____

Answer 1

Answer:

55 / 2048

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Step-by-step explanation:

The probability of getting k tails in n tosses

= P(k tails)

= ( # ways of choosing k positiions out of n to be the tails ) / ( # total number of possible outcomes )

= $\binom{n}{k} / 2^n$

So

P(4 tails) = P(7 tails)

=> $\binom{n}{4} = \binom{n}{7}$

=> n = 4+7 = 11

Therefore

P(2 tails)

= $\binom{11}{2} / 2^{11}$

= 55 / 2048

Answer 2

Answer:

Concept:

Probability is the branch of mathematics that deals with numerical descriptions of how likely an event is to occur or how likely a proposition is to be true.

Step-by-step explanation:

$\begin{gathered}x=\text { no. of heads. } \\P(x=r)={ }^{n} C_{r} p^{r} q^{n-r} . \\p=p r o b \text { of getting head }=\frac{1}{2} \\q=1-\frac{1}{2}=\frac{1}{2} . \\P(x=4)=p(x=7)\end{gathered}$

${ }^{n} C_{4}\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{n-4}={ }^{n} C_{7}\left(\frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)^{n-7}$

${ }^{n} C_{4}\left(\frac{1}{2}\right)^{n}={ }^{n} C_{7}\left(\frac{1}{2}\right)^{n}$

${ }^{n} C_{4}={ }^{n} C_{7}$

n = 4 + 7

n = 11

Probability of getting 2 tails are

$\begin{aligned}P(x=2) &={ }^{11} C_{2}\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{9} \\&=\frac{11 \times 10}{2} \times \frac{1}{(2)^{11}}\end{aligned}$

                = $\frac{55}{2048}$

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