Three different coins are tossed together find the probability of getting exactly 1 tail

Answer

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:

Hence, we get the all required probabilities.

Note:

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Given: Three coins are tossed simultaneously.

When three coins are tossed then the outcome will be

TTT, THT, TTH, THH. HTT, HHT, HTH, HHH.

Hence total number of outcome is 8.

(i) For exactly two head we get favorable outcome as THH, HHT ,HTH

So, total number of favorable outcome i.e. exactly two head 3

We know that;

Probability = $\frac{Numberoffavorableevent}{Totalnumberofevent}$

Hence probability of getting exactly two head is $\frac{3}{8}$

(ii) In case, at least two head we have favorable outcome as HHT, HTH, HHH ,THH

So, total number of favorable outcome i.e. at least two head is 4

We know that;

Probability = $\frac{Numberoffavorableevent}{Totalnumberofevent}$

Hence, probability of getting at least two head when three coins are tossed simultaneously is equal to $\frac{4}{8}$ = $\frac{1}{2}$

(iiii) At least one head and one tail we get in case THT, TTH, THH. HTT, HHT, HTH,

So, total number of favorable outcome i.e. at least one tail and one head is 6

We know that;

Probability = $\frac{Numberoffavorableevent}{Totalnumberofevent}$

Hence, probability of getting at least one head and one tail is equal to $\frac{6}{8}$ = $\frac{3}{4}$

(iv) No tail i.e HHH

Hence, total number of favorable outcome is 1

We know that;

Probability = $\frac{Numberoffavorableevent}{Totalnumberofevent}$

Hence, probability of getting no tails is $\frac{1}{8}$

Mathematics

RD Sharma

Standard X