Question

Roll a 6-sided die and a 10-sided die. (Both dice are fair and have an ace side.) What is the chance that they will both land on ace? 1/60 (Give exact answer.) What is the chance that neither will land on ace? 3/4 (Give exact answer.) What is the chance that at least one will not land on ace?

Answer 1

Answer:

a.  $P(Both) = \frac{1}{60}$

b.   $P(None) = \frac{3}{4}$

c.  $P(At least 1) = \frac{1}{4}$

Step-by-step explanation:

Given

A fair 6-sided die

A fair 10-sided die

Both have an ace side

For the 6-sided die;

Probability of landing on Ace: P(A)

$P(A) = \frac{1}{6}$

Probability of not landing on Ace: P(A')

$P(A') = \frac{5}{6}$

For the 10-sided die;

Probability of landing on Ace: P(B)

$P(B) = \frac{1}{10}$

Probability of not landing on Ace: P(B')

$P(B') = \frac{9}{10}$

Solving (a): Both landing on Ace

This is calculated as thus;

$P(Both) = P(A) * P(B)$

$P(Both) = \frac{1}{6} * \frac{1}{10}$

$P(Both) = \frac{1}{60}$

Solving (b): None landing on Ace

This is calculated as thus;

$P(None) = P(A') * P(B')$

$P(None) = \frac{5}{6} * \frac{9}{10}$

$P(None) = \frac{1}{6} * \frac{9}{2}$

$P(None) = \frac{1}{2} * \frac{3}{2}$

$P(None) = \frac{3}{4}$

Solving (c): At least one landing on Ace

This is calculated as thus;

$P(At least 1) = 1 - P(None)$

$P(At least 1) = 1 - \frac{3}{4}$

$P(At least 1) = \frac{4 - 3}{4}$

$P(At least 1) = \frac{1}{4}$

Answer 2

The required probabilities are,

(a):$P(A)=\frac{1}{60}$

(b):$P(B)=\frac{3}{4}$

(c):$P(C)=\frac{59}{60}$

Given that the die having the 6 sides.

The formula for finding the expected probability is,

$P(A)=\frac{m}{n}$

Where m= Number of expected observation

n= Number of total observation

Part(a):

P(Both landed on ace) is,

$P(A)=\frac{1}{6} \times\frac{1}{10} \\=\frac{1}{60}$

Part(b):

P(Neither landed on ace) is,

$P(B)=[1-\frac{1}{6} ][1-\frac{1}{10} ]\\=\frac{3}{4}$

Part(c):

P(At least one not land on ace) = 1 - P(Both landed on ace)

$1-\frac{1}{60} =\frac{59}{60}$

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