Question

Two fair coins are tossed. What is the probability of getting at the most one head?​

Answer 1

HINT:

First, find the sample space of the problem. Next, find the number of elements in the sample space = total number of outcomes. Now, let E be an event when the head appears on one coin and the tail on another coin. Find the set E and find the number of elements in the sample space = total number of outcomes. Next, use the formula.

Complete step-by-step answer:

In this question, we are given that two fair coins are tossed simultaneously.We need to find the probability of getting Head on one coin and Tail on the other coin is

n (S) = 4

E be an event when the head appears on one coin and the tail on another coin.

E:{H,T,T,H}

So, the number of elements in the

E = number of favourable outcomes is given by the following:

n (E) = 2

Now, we know that probability of an event is equal to the number of favourable outcomes divided by the total number of outcomes. i.e.

$\\ \\ \\ P(E)= \frac{number \: of \: favourable \: outcomes}{total \: number \: of \: outcomes}$

$Using \: the \: above \: formula, \\ \: we \: will \: get \: the following: \\ \\ </em></p><p></p><p><em>[tex] P(E) = \frac{2}{4} = \frac{1}{2} \: \: \: answer \\ \\ \\ so \: \\ \\ \frac{1}{2} \: is \: \: the \: probability \: of \: getting \: \\ Head \: on \: one \: coin$

Answer 2

Answer:

then

\: \: \: \: (1) \: tan \theta \: = \: \sqrt{ {r}^{2} - 1}(1)tanθ=

r

2

−1

\: \: \: \: (2) \: cos \theta \: = \: r(2)cosθ=r

\: \: \: \: (3) \:sin \theta \: + cos \theta \: = \: \dfrac{ \sqrt{1 + {r}^{2} } }{r}(3)sinθ+cosθ=

r

1+r

2

\: \: \: \: (4) \: cot \theta \: = \: \sqrt{1 - {r}^{2}}(4)cotθ=

1−r

2

then

\: \: \: \: (1) \: tan \theta \: = \: \sqrt{ {r}^{2} - 1}(1)tanθ=

r

2

−1

\: \: \: \: (2) \: cos \theta \: = \: r(2)cosθ=r

\: \: \: \: (3) \:sin \theta \: + cos \theta \: = \: \dfrac{ \sqrt{1 + {r}^{2} } }{r}(3)sinθ+cosθ=

r

1+r

2

\: \: \: \: (4) \: cot \theta \: = \: \sqrt{1 - {r}^{2}}(4)cotθ=

1−r

2