Three coins are tossed together. Find the probability of getting:

Question:

Three coins are tossed together. Find the probability of getting:

(i) exactly two heads

(ii) at least two heads

(iii) at least one head and one tail

(iv) no tails

Solution:

When 3 coins are tossed together, the outcomes are as follows:

$\mathrm{S}=\{(\mathrm{h}, \mathrm{h}, \mathrm{h}),(\mathrm{h}, \mathrm{h}, \mathrm{t}),(\mathrm{h}, \mathrm{t}, \mathrm{h}),(\mathrm{h}, \mathrm{t}, \mathrm{t}),(\mathrm{t}, \mathrm{h}, \mathrm{h}),(\mathrm{t}, \mathrm{h}, \mathrm{t}),(\mathrm{t}, \mathrm{t}, \mathrm{h}),(\mathrm{t}, \mathrm{t}, \mathrm{t})\}$

Therefore, the total number of outcomes is 8 .

(i) Let A be the event of getting triplets having exactly 2 heads.

Triplets having exactly 2 heads: $(\mathrm{h}, \mathrm{h}, \mathrm{t}),(\mathrm{h}, \mathrm{t}, \mathrm{h}),(\mathrm{t}, \mathrm{h}, \mathrm{h})$

Therefore, the total number of favourable outcomes is 3 .

$\mathrm{P}(A)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{8}$

(ii) Let $A$ be the event of getting triplets having at least 2 heads.

Triplets having at least 2 heads: $(\mathrm{h}, \mathrm{h}, \mathrm{t}),(\mathrm{h}, \mathrm{t}, \mathrm{h}),(\mathrm{t}, \mathrm{h}, \mathrm{h}),(\mathrm{h}, \mathrm{h}, \mathrm{h})$

Therefore, the total number of favourable outcomes is 4 .

$\mathrm{P}(A)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{4}{8}=\frac{1}{2}$

(iii) Let A be the event of getting triplets having at least one head and one tail.

Triplets having at least one head and one tail: $(\mathrm{h}, \mathrm{h}, \mathrm{t}),(\mathrm{h}, \mathrm{t}, \mathrm{h}),(\mathrm{t}, \mathrm{h}, \mathrm{h}),(\mathrm{h}, \mathrm{h}, \mathrm{t}),(\mathrm{t}, \mathrm{t}, \mathrm{h}),(\mathrm{t}, \mathrm{h}, \mathrm{t})$

Therefore, the total number of favourable outcomes is 6 .

$\mathrm{P}(A)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{6}{8}=\frac{3}{4}$

(iv) Let $A$ be the event of getting triplets having no tail.

Triplets having no tail: $(\mathrm{h}, \mathrm{h}, \mathrm{h})$

Therefore, the total number of favourable outcomes is $1 .$

$\mathrm{P}(A)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{1}{8}$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now