Question

A fair coin is tossed $6$ times. What is the probability of getting at least $3$ heads?
A) $\dfrac{{11}}{{16}}$
B) $\dfrac{{21}}{{32}}$
C) $\dfrac{1}{{18}}$
D) $\dfrac{3}{{64}}$

Answer 1

Hint: Probability can be defined as the state of being probable and which is at the extent to which something is likely to happen in the particular situations or the favourable outcomes. Probability of any given event is given by the ratio of the favourable outcomes with the total number of the outcomes. Here we will use the frame formula to get at least three heads.

Complete step by step solution:
Let us assume that “X” denotes the number of heads when a coin is tossed six times.
Now, for binomial distribution
$n = 6$
Now the probability of getting head is $p = \dfrac{1}{2}$
And the probability of getting not head or tail is $q = \dfrac{1}{2}$
Probability is given by - $P(X = r) = {}^6{C_r}{\left( {\dfrac{1}{2}} \right)^r}{\left( {\dfrac{1}{2}} \right)^{6 - r}}$
Simplification implies
$P(X = r) = {}^6{C_r}{\left( {\dfrac{1}{2}} \right)^6}$
Here, $r = 0,1,2,....6$
Now, the required probability of getting ahead at least three times.
\[P(X \geqslant 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)\]
Place formula in the above expression
\[P(X \geqslant 3) = {}^6{C_3}{\left( {\dfrac{1}{2}} \right)^6} + {}^6{C_4}{\left( {\dfrac{1}{2}} \right)^6} + {}^6{C_5}{\left( {\dfrac{1}{2}} \right)^6} + {}^6{C_6}{\left( {\dfrac{1}{2}} \right)^6}\]
Take common multiple common from the above expression –
\[P(X \geqslant 3) = {\left( {\dfrac{1}{2}} \right)^6}[{}^6{C_3} + {}^6{C_4} + {}^6{C_5} + {}^6{C_6}]\]
Simplify the above expression –
\[P(X \geqslant 3) = {\left( {\dfrac{1}{2}} \right)^6}[20 + 15 + 6 + 1]\]
Simplify the above expression finding the sum of the terms in the above expression.
\[P(X \geqslant 3) = \left( {\dfrac{1}{{64}}} \right)[42]\]
Common multiples from the numerator and the denominator cancel each other.
\[P(X \geqslant 3) = \left( {\dfrac{{21}}{{32}}} \right)\]
Therefore, option (B) is the correct answer.

Note:
Always remember that the probability of any event always ranges between zero and one. It can never be the negative number or the number which is greater than one. The probability of impossible events is always stated equal to zero whereas, the probability of the sure event is always stated equal to one.