Question

Answer only those who thinks that probability is the most easy chapter.

Roushi has the following coins in her pocket: ₹1, ₹2, ₹5, ₹10. She selects one coin at random to puts in a charity collection box. What is the probability that she
(i) gives more than 1?
(ii) has less than 15 left in her pocket?​

Answer 1

Given Question

Roushi has the following coins in her pocket: ₹1, ₹2, ₹5, ₹10. She selects one coin at random to puts in a charity collection box. What is the probability that she

(i) gives more than 1?

(ii) has less than 15 left in her pocket?

$\large\underline{\sf{Solution-}}$

Given that

• Roushi has the following coins in her pocket: ₹1, ₹2, ₹5, ₹10.

• She selects one coin at random to puts in a charity collection box.

So, possible outcomes are {1, 2, 5, 10}

It means,

Number of elements in Sample space, n(S) = 4

(i) Now, we have to find the probability that she give more than ₹ 1 coin.

Let assume that

E = { getting more than ₹ 1 coin }

It means,

Favourable outcomes (E) = {2, 5, 10}

It means,

Number of elements in Favourable outcomes, n(E) = 3

We know, Probability of an event, P(E) is given by

$\red{\rm :\longmapsto\:\boxed{\tt{ P(E) = \frac{n(E)}{n(S)} }}}$

So, on substituting the values, we get

$\green{\rm :\longmapsto\:\boxed{\tt{ P(E) = \frac{3}{4} }}}$

(ii) Now, we have to find the probability that she has less than 15 left in her pocket?.

Let assume that

E = { getting ₹ 15 less in her pocket }

↝ If she pick ₹ 1 coin, she left with ₹ 17 in her pocket.

↝ If she pick ₹ 2 coin, she left with ₹ 16 in her pocket.

↝ If she pick ₹ 5 coin, she left with ₹ 13 in her pocket.

↝ If she pick ₹ 10 coin, she left with ₹ 8 in her pocket.

It means,

Favourable outcomes (E) = {5, 10}

It means,

Number of elements in Favourable outcomes, n(E) = 2

So, required probability of an event is

$\green{\rm :\longmapsto\:\boxed{\tt{ P(E) = \frac{2}{4} = \frac{1}{2} }}}$

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Explore more :-

The sample space of a random experiment is collection of all possible outcomes.

The probability of any outcome is a number between 0 and 1 including 0 and 1.

The probability of sure event is 1.

The probability of impossible event is 0.

P(E) + P(E') = 1