Probability Part 2

Coin is tossed 5 times, so n=5
Let, p= probability of getting heads
q= probability of getting tails.
∴p+q=1......(1)
∴ Probability of getting 4 heads
=‌5C4⋅p4⋅q
And probability of getting 5 heads
=‌5C5⋅p5
Given, ‌5C4⋅p4⋅q=‌5C5⋅p5
⇒5q=p......‌ (2) ‌
From equation (1) and (2), we get,
5q+q=1
⇒6q=1

⇒q=‌

1

6

∴p=1−‌

1

6

=‌

5

6

Now, probability of getting atmost two heads
=p(x=0)+p(x=1)+p(x=2)
p(x=0)= Getting zero head in 5 trials
=‌5C0⋅p0⋅q5
p(x=1)= Getting one head in 5 trials
=‌5C1⋅p1⋅q4
p(x=2)= Getting two heads in 5 trials
=‌5C2⋅p2⋅q3
=‌5C0⋅q5+‌5C1⋅pq4+‌5C2⋅p2q3
=(‌

1

6

)5+5⋅‌

5

6

⋅(‌

1

6

)4+10⋅(‌

5

6

)2⋅(‌

1

6

)3
=‌

1+25+250

65

=‌

276

65

=‌

46

64