Question

Three different coins are tossed together. Find the probability of:
(i)Exactly two heads
(ii) At least one tail
(iii) At most two heads
(iv) At most two tails
(v) At least two tails
(vi) At most one head
(vii) At least one head

Answer 1

Hint: First, before proceeding for this, we must know that we require the total number of cases when three different coins are tossed which are 8. Then, we will find the favourable outcomes for all the cases that are asked in the question. Then, after getting that we will get the probabilities of all the cases as required.

Complete step by step answer:
In this question, we are supposed to find the probabilities of different conditions as given in the question when three different coins are tossed together.
So, before proceeding for this, we must know that we requires the total number of cases when three different coins are tossed as:
(HHH, HHT, THH, HTH, TTH, THT, HTT, TTT)
So, we get the total number of cases when three different coins are tossed as 8.
Now, for the case (i), we are asked to find the probability when we get exactly two heads.
Then, we get the favourable cases as:
(HHT, THH, HTH) which are 3 in number.
So, we get the probability of exactly two heads in 3 throws of different coins as:
$\dfrac{3}{8}$
Now, for the case (ii), we are asked to find the probability when we get at least one tail.
Then, we get the favourable cases as:
(HHT, THH, HTH, TTH, TTT, THT, HTT) which are 7 in number.
So, we get the probability of at least one tail in 3 throws of different coins as:
$\dfrac{7}{8}$
Now, for the case (iii), we are asked to find the probability when we get at most two heads.
Then, we get the favourable cases as:
(HHT, THH, HTH, TTH, TTT, THT, HTT) which are 7 in number.
So, we get the probability of at most two heads in 3 throws of different coins as:
$\dfrac{7}{8}$
Now, for the case (iv), we are asked to find the probability when we get at most two tails.
Then, we get the favourable cases as:
(HHT, THH, HTH, HHH, TTH, THT, HTT) which are 7 in number.
So, we get the probability of at most two tails in 3 throws of different coins as:
$\dfrac{7}{8}$
Now, for the case (v), we are asked to find the probability when we get at least two tails.
Then, we get the favourable cases as:
(HTT, THT, TTH, TTT) which are 4 in number.
So, we get the probability of at least two tails in 3 throws of different coins as:
$\dfrac{4}{8}$
Now, for the case (vi), we are asked to find the probability when we get at most one head.
Then, we get the favourable cases as:
(HTT, THT, TTH, TTT) which are 4 in number.
So, we get the probability of at most one head in 3 throws of different coins as:
$\dfrac{4}{8}$
Now, for the case (vii), we are asked to find the probability when we get at least one head.
Then, we get the favourable cases as:
(HHT, THH, HTH, TTH, THT, HTT, TTT) which are 7 in number.
So, we get the probability of at least one head in 3 throws of different coins as:
$\dfrac{7}{8}$
Hence, we get the all required probabilities.

Note:
Now, to solve these types of questions we need to know some of the basics of probability in advance to proceed in these types of questions. So, probability is defined as the ratio of the favourable outcomes to the total outcomes and it is always in the form of ratio.