Question

When three unbiased coins are tossed together, what is the probability of not getting two tails and one head in any order?

a) 1/8
b) 3/8
c) 5/8
d) 7/8​

Answer 1

Answer:

ANSWER

Here S=( TTT,TTH,THT,HTT,THH,HTH,HHT,HHH )

a(i) one tail (THH,HTH,HHH)

Then E=3

So (p)=

n(S)

n(E)

=

8

3

a(ii) two tails

Then E=3

So (p)=

n(S)

n(E)

=

8

3

a(iii) all tails

Then E=1

So (p)=

n(S)

n(E)

=

8

1

a(iv) at least two tails

Then E=4

So (p)=

n(S)

n(E)

=

8

4

=

2

1

(b) (i) at most two tails

Then E=7

So (p)=

n(S)

n(E)

=

8

7

b(ii) At most two heads

Then E=7

So (p)=

n(S)

n(E)

=

8

7

Answer 2

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Given:

Three unbiased coins are tossed together

To find:

Probability of not getting two tails and

one head = ?

Solution:

Let S be the sample space of tossing three unbiased coins

T = tails

H = head

S = { TTT, TTH, THH, THT, HHH, HHT, HTT, HTH }

n(S) = 8

Let A be the event of not getting two tails and one head in any order

A = { TTT, THH, HHH, HHT, HTH }

n(A) = 5

$\\ \\ P(A) = \dfrac{n(A)}{n(S)} \\ \\ = \dfrac{5}{8}$

Answer:

Thus, probability of not getting two tails and one head in any order is 5/8.

So, Correct answer is option C = 5/8

Knowledge booster:

☞ Probability of an event of finite sample space S is written as P(A) and is defined as

$\\ \\ P(A) = \\ \\ \dfrac{number \: of \: sample \: points \: in \: A}{number \: of \: sample \: points \: in \: S} \\ \\ = \dfrac {n(A)}{n(S)}$

☞ The set of all possible outcomes of random experiment is called sample space and denoted by S or omega

☞ Try to solve more such questions to get good hold on it